Algebra - Algebraic Expressions

Simplify: $\;$ $\dfrac{\left(a^{\frac{1}{m}} - a^{\frac{1}{n}}\right)^2 + 4 \; a^{\frac{m + n}{mn}}}{\left(a^{\frac{2}{m}} - a^{\frac{2}{n}}\right) \left(\sqrt[m]{a^{m + 1}} + \sqrt[n]{a^{n + 1}}\right)}$

$\dfrac{\left(a^{\frac{1}{m}} - a^{\frac{1}{n}}\right)^2 + 4 \; a^{\frac{m + n}{mn}}}{\left(a^{\frac{2}{m}} - a^{\frac{2}{n}}\right) \left(\sqrt[m]{a^{m + 1}} + \sqrt[n]{a^{n + 1}}\right)}$

$= \dfrac{a^{\frac{2}{m}} + a^{\frac{2}{n}} - 2 \; a^{\frac{1}{m}} \; a^{\frac{1}{n}} + 4 \; a^{\frac{1}{m} + \frac{1}{n}}}{\left(a^{\frac{2}{m}} - a^{\frac{2}{n}}\right) \left(a^{\frac{m + 1}{m}} + a^{\frac{n + 1}{n}}\right)}$

$= \dfrac{a^{\frac{2}{m}} + a^{\frac{2}{n}} + 2 \; a^{\frac{1}{m} + \frac{1}{n}}}{\left[\left(a^{\frac{1}{m}}\right)^2 - \left(a^{\frac{1}{n}}\right)^2\right] \left(a^{\frac{m + 1}{m}} + a^{\frac{n + 1}{n}}\right)}$

$= \dfrac{\left(a^{\frac{1}{m}} + a^{\frac{1}{n}}\right)^2}{\left(a^{\frac{1}{m}} + a^{\frac{1}{n}}\right) \left(a^{\frac{1}{m}} - a^{\frac{1}{n}}\right) \left(a^1 \times a^{\frac{1}{m}} + a^1 \times a^{\frac{1}{n}}\right)}$

$= \dfrac{a^{\frac{1}{m}} + a^{\frac{1}{n}}}{\left(a^{\frac{1}{m}} - a^{\frac{1}{n}}\right) \times a \times \left(a^{\frac{1}{m}} + a^{\frac{1}{n}}\right)}$

$= \dfrac{1}{a \left(a^{\frac{1}{m}} - a^{\frac{1}{n}}\right)}$

Algebra - Algebraic Expressions

Simplify: $\;$ $\left(\dfrac{a + b}{a - b}\right)^{\frac{1}{2}} - \dfrac{2a \sqrt{a^2 - b^2}}{b^2 \left(ab^{-1} + 1\right)^2} \times\dfrac{1}{1 + \dfrac{1 - b a^{-1}}{1 + ba^{-1}}}$

$\left(\dfrac{a + b}{a - b}\right)^{\frac{1}{2}} - \dfrac{2a \sqrt{a^2 - b^2}}{b^2 \left(ab^{-1} + 1\right)^2} \times\dfrac{1}{1 + \dfrac{1 - b a^{-1}}{1 + ba^{-1}}}$

$= \dfrac{\left(\sqrt{a + b}\right) \left(\sqrt{a + b}\right)}{\left(\sqrt{a - b}\right) \left(\sqrt{a + b}\right)} - \dfrac{2a \sqrt{a^2 - b^2}}{b^2 \left(\dfrac{a}{b} + 1\right)^2} \times \dfrac{1}{1 + \dfrac{1 - \dfrac{b}{a}}{1 + \dfrac{b}{a}}}$

$= \dfrac{a + b}{\sqrt{a^2 - b^2}} - \dfrac{2a \sqrt{a^2 - b^2}}{\left(a + b\right)^2} \times \dfrac{1}{1 + \dfrac{a - b}{a + b}}$

$= \dfrac{a + b}{\sqrt{a^2 - b^2}} - \dfrac{2a \sqrt{a^2 - b^2}}{\left(a + b\right)^2} \times \dfrac{a + b}{a + b + a - b}$

$= \dfrac{a + b}{\sqrt{a^2 - b^2}} - \dfrac{2a \sqrt{a^2 - b^2}}{a + b} \times \dfrac{1}{2a}$

$= \dfrac{a + b}{\sqrt{a^2 - b^2}} - \dfrac{\sqrt{a^2 - b^2}}{a + b}$

$= \dfrac{\left(a + b\right)^2 - \left(a^2 - b^2\right)}{\left(a + b\right) \sqrt{a^2 - b^2}}$

$= \dfrac{a^2 + b^2 + 2ab - a^2 + b^2}{\left(a + b\right) \sqrt{a^2 - b^2}}$

$= \dfrac{2b^2 + 2ab}{\left(a + b\right) \sqrt{a^2 - b^2}}$

$= \dfrac{2b \left(a + b\right)}{\left(a + b\right) \sqrt{a^2 - b^2}}$

$= \dfrac{2b}{\sqrt{a^2 - b^2}}$

Algebra - Algebraic Expressions

Simplify: $\;$ $\left[\dfrac{a^3 - 8}{a^2 - 5a + 6} - \dfrac{\left(a + 1\right)^2 + 3}{a - 3} + \dfrac{a^2 + a}{\sqrt[4]{a}}\right] : \dfrac{\sqrt{ab}}{\sqrt[4]{a^{-1} b^2}}$

$\left[\dfrac{a^3 - 8}{a^2 - 5a + 6} - \dfrac{\left(a + 1\right)^2 + 3}{a - 3} + \dfrac{a^2 + a}{\sqrt[4]{a}}\right] : \dfrac{\sqrt{ab}}{\sqrt[4]{a^{-1} b^2}}$

$= \left[\dfrac{\left(a - 2\right) \left(a^2 + 2a + 4\right)}{a^2 - 3a - 2a + 6} - \dfrac{a^2 + 2a + 1 + 3}{a - 3} + \dfrac{a \left(a + 1\right)}{a^{\frac{1}{4}}} \right] : \dfrac{a^{\frac{1}{2}} b^{\frac{1}{2}}}{\left(a^{-1}\right)^{\frac{1}{4}} \left(b^2\right)^{\frac{1}{4}}}$

$= \left[\dfrac{\left(a - 2\right) \left(a^2 + 2a + 4\right)}{a \left(a - 3\right) - 2 \left(a - 3\right)} - \dfrac{a^2 + 2a + 4}{a - 3} + a^{\frac{3}{4}} \left(a + 1\right)\right] : a^{\frac{1}{2}} b^{\frac{1}{2}} a^{\frac{1}{4}} b^{\frac{-1}{2}}$

$= \left[\dfrac{\left(a - 2\right) \left(a^2 + 2a + 4\right)}{\left(a - 2\right) \left(a - 3\right)} - \dfrac{a^2 + 2a + 4}{a - 3} + a^{\frac{3}{4}} \left(a + 1\right)\right] : a^{\frac{3}{4}}$

$= \left[\dfrac{a^2 + 2a + 4 - a^2 - 2a - 4}{a - 3} + a^{\frac{3}{4}} \left(a + 1\right)\right] : a^{\frac{3}{4}}$

$= \dfrac{0 + a^{\frac{3}{4}} \left(a + 1\right)}{a^{\frac{3}{4}}}$

$= a + 1$

Algebra - Algebraic Expressions

Simplify: $\;$ $\left(a^2 \sqrt{b}\right)^{\frac{-1}{2}} \left(\sqrt{ab} - \dfrac{ab}{a + \sqrt{ab}}\right) : \dfrac{\sqrt[4]{ab} - \sqrt{b}}{a - b}$

$\left(a^2 \sqrt{b}\right)^{\frac{-1}{2}} \left(\sqrt{ab} - \dfrac{ab}{a + \sqrt{ab}}\right) : \dfrac{\sqrt[4]{ab} - \sqrt{b}}{a - b}$

$= \left(\dfrac{1}{\left(a^2\right)^{\frac{1}{2}} \left(b^{\frac{1}{2}}\right)^{\frac{1}{2}}} \times \dfrac{a \sqrt{ab} + ab - ab}{a + \sqrt{ab}} \right) : \dfrac{\sqrt[4]{ab} - \sqrt{b}}{a - b}$

$= \dfrac{1}{a b^{\frac{1}{4}}} \times \dfrac{a \times a^{\frac{1}{2}} b^{\frac{1}{2}}}{a + \sqrt{ab}} \times \dfrac{a - b}{\sqrt[4]{ab} - \sqrt{b}}$

$= \dfrac{a^{\frac{1}{2}} b^{\frac{1}{4}}}{a + a^{\frac{1}{2}} b^{\frac{1}{2}}} \times \dfrac{a - b}{a^{\frac{1}{4}} b^{\frac{1}{4}} - b^{\frac{1}{2}}}$

$= \dfrac{a^{\frac{1}{2}} b^{\frac{1}{4}}}{a^{\frac{1}{2}} \left(a^{\frac{1}{2}} + b^{\frac{1}{2}}\right)} \times \dfrac{a - b}{b^{\frac{1}{4}} \left(a^{\frac{1}{4}} - b^{\frac{1}{4}}\right)}$

$= \dfrac{a - b}{\left(a^{\frac{1}{2}} + b^{\frac{1}{2}}\right) \left(a^{\frac{1}{4}} - b^{\frac{1}{4}}\right)}$

$= \dfrac{\left(a - b\right) \left(a^{\frac{1}{4}} + b^{\frac{1}{4}}\right)}{\left(a^{\frac{1}{2}} + b^{\frac{1}{2}}\right) \left(a^{\frac{1}{4}} - b^{\frac{1}{4}}\right) \left(a^{\frac{1}{4}} + b^{\frac{1}{4}}\right)}$

$= \dfrac{\left(a - b\right) \left(a^{\frac{1}{4}} + b^{\frac{1}{4}}\right)}{\left(a^{\frac{1}{2}} + b^{\frac{1}{2}}\right) \left(a^{\frac{1}{2}} - b^{\frac{1}{2}}\right)}$

$= \dfrac{\left(a - b\right) \left(a^{\frac{1}{4}} + b^{\frac{1}{4}}\right)}{a - b}$

$= a^{\frac{1}{4}} + b^{\frac{1}{4}}$

Algebra - Algebraic Expressions

Simplify: $\;$ $\left(\dfrac{a \sqrt{a} + b \sqrt{b}}{\sqrt{a} + \sqrt{b}}\right) : \left(a - b\right) + \dfrac{2 \sqrt{b}}{\sqrt{a} + \sqrt{b}}$

$\left(\dfrac{a \sqrt{a} + b \sqrt{b}}{\sqrt{a} + \sqrt{b}}\right) : \left(a - b\right) + \dfrac{2 \sqrt{b}}{\sqrt{a} + \sqrt{b}}$

$= \dfrac{\left(a \sqrt{a} + b \sqrt{b}\right) \left(\sqrt{a} - \sqrt{b}\right)}{\left(\sqrt{a} + \sqrt{b}\right) \left(\sqrt{a} - \sqrt{b}\right)} \times \dfrac{1}{a - b} + \dfrac{2 \sqrt{b} \left(\sqrt{a} - \sqrt{b}\right)}{\left(\sqrt{a} + \sqrt{b}\right) \left(\sqrt{a} - \sqrt{b}\right)}$

$= \dfrac{\left(a^2 - a \sqrt{ab} + b \sqrt{ab} - b^2\right)}{a - b} \times \dfrac{1}{a - b} + \dfrac{2 \sqrt{b} \left(\sqrt{a} - \sqrt{b}\right)}{a - b}$

$= \dfrac{\left(a + b\right) \left(a - b\right) - \sqrt{ab} \left(a - b\right)}{\left(a - b\right)^2} + \dfrac{2 \sqrt{b} \left(\sqrt{a} - \sqrt{b}\right)}{den}$

$= \dfrac{\left(a - b\right) \left(a + b - \sqrt{ab}\right)}{\left(a - b\right)^2} + \dfrac{2 \sqrt{ab} - 2b}{a - b}$

$= \dfrac{a + b - \sqrt{ab}}{a - b} + \dfrac{2 \sqrt{ab} - 2b}{a - b}$

$= \dfrac{a - b + \sqrt{ab}}{a - b}$

$= 1 + \dfrac{\sqrt{ab}}{a - b}$

Algebra - Algebraic Expressions

Simplify: $\;$ $\left[\left(\sqrt[4]{a} - \sqrt[4]{b}\right)^{-2} + \left(\sqrt[4]{a} + \sqrt[4]{b}\right)^{-2}\right] : \left(\dfrac{\sqrt{a} + \sqrt{b}}{a - b}\right)^2$

$\left[\left(\sqrt[4]{a} - \sqrt[4]{b}\right)^{-2} + \left(\sqrt[4]{a} + \sqrt[4]{b}\right)^{-2}\right] : \left(\dfrac{\sqrt{a} + \sqrt{b}}{a - b}\right)^2$

$= \left[\dfrac{1}{\left(a^{\frac{1}{4}} - b^{\frac{1}{4}}\right)^2} + \dfrac{1}{\left(a^{\frac{1}{4} } + b^{\frac{1}{4}}\right)^2}\right] : \left(\dfrac{\sqrt{a} + \sqrt{b}}{a - b}\right)^2$

$= \dfrac{\dfrac{a^{\frac{1}{2}} + b^{\frac{1}{2}} + 2 a^{\frac{1}{4}} b^{\frac{1}{4}} + a^{\frac{1}{2}} + b^{\frac{1}{2}} - 2 a^{\frac{1}{4}} b^{\frac{1}{4}}}{\left(a^{\frac{1}{4}} - b^{\frac{1}{4}}\right)^2 \left(a^{\frac{1}{4}} + b^{\frac{1}{4}}\right)^2}}{\left(\dfrac{\sqrt{a} + \sqrt{b}}{a - b}\right)^2}$

$= \dfrac{2 \left(\sqrt{a} + \sqrt{b}\right)}{\left(a^{\frac{1}{2}} - b^{\frac{1}{2}}\right)^2} \times \dfrac{\left(a - b\right)^2}{\left(\sqrt{a} + \sqrt{b}\right)^2}$

$= \left[\dfrac{2}{\sqrt{a} + \sqrt{b}}\right] \times \left[\dfrac{a - b}{\sqrt{a} - \sqrt{b}}\right]^2$

$= \left[\dfrac{2}{\sqrt{a} + \sqrt{b}}\right] \times \left[\dfrac{\left(a - b\right) \left(\sqrt{a} + \sqrt{b}\right)}{\left(\sqrt{a} - \sqrt{b}\right) \left(\sqrt{a} + \sqrt{b}\right)}\right]^2$

$= \left[\dfrac{2}{\sqrt{a} + \sqrt{b}}\right] \times \left[\dfrac{\left(a - b\right) \left(\sqrt{a} + \sqrt{b}\right)}{a - b}\right]^2$

$= 2 \left(\sqrt{a} + \sqrt{b}\right)$

Algebra - Algebraic Expressions

Simplify: $\;$ $\dfrac{2 \sqrt{b}}{\sqrt{a} + \sqrt{b}} + \left\{\dfrac{a^{\frac{3}{2}} + b^{\frac{3}{2}}}{\sqrt{a} + \sqrt{b}} - \dfrac{1}{\left(ab\right)^{\frac{1}{2}}}\right\} \left(a - b\right)^{-1}$

$\dfrac{2 \sqrt{b}}{\sqrt{a} + \sqrt{b}} + \left\{\dfrac{a^{\frac{3}{2}} + b^{\frac{3}{2}}}{\sqrt{a} + \sqrt{b}} - \dfrac{1}{\left(ab\right)^{\frac{1}{2}}}\right\} \left(a - b\right)^{-1}$

$= \dfrac{2 \sqrt{b}}{\sqrt{a} + \sqrt{b}} + \left\{\dfrac{\left(a^{\frac{1}{2}}\right)^3 + \left(b^{\frac{1}{2}}\right)^3}{\sqrt{a} + \sqrt{b}} - \left(ab\right)^{\frac{1}{2}} \right\} \times \dfrac{1}{\left(a - b\right)}$

$= \dfrac{2 \sqrt{b}}{\sqrt{a} + \sqrt{b}} + \left\{\dfrac{\left(\sqrt{a} + \sqrt{b}\right) \left[\left(\sqrt{a}\right)^2 - \sqrt{a} \sqrt{b} + \left(\sqrt{b}\right)^2\right]}{\sqrt{a} + \sqrt{b}} - \sqrt{ab} \right\} \times \dfrac{1}{\left(a - b\right)}$

$= \dfrac{2 \sqrt{b}}{\sqrt{a} + \sqrt{b}} + \dfrac{a - \sqrt{ab} + b - \sqrt{ab}}{a - b}$

$= \dfrac{2 \sqrt{b}}{\sqrt{a} + \sqrt{b}} + \dfrac{a - 2 \sqrt{ab} + b}{a - b}$

$= \dfrac{2 \sqrt{b}}{\sqrt{a} + \sqrt{b}} + \dfrac{\left(\sqrt{a} - \sqrt{b}\right)^2}{a - b}$

$= \dfrac{2 \sqrt{b}}{\sqrt{a} + \sqrt{b}} + \dfrac{\left(\sqrt{a} - \sqrt{b}\right)^2}{\left(\sqrt{a} + \sqrt{b}\right) \left(\sqrt{a} - \sqrt{b}\right)}$

$= \dfrac{2 \sqrt{b}}{\sqrt{a} + \sqrt{b}} + \dfrac{\sqrt{a} - \sqrt{b}}{\sqrt{a} + \sqrt{b}}$

$= \dfrac{\sqrt{a} + \sqrt{b}}{\sqrt{a} + \sqrt{b}}$

$= 1$

Algebra - Algebraic Expressions

Simplify: $\;$ $\left(a + a^{\frac{1}{2}} b^{\frac{1}{2}}\right) \left(a + b\right)^{-1} \left[\sqrt{a} \left(\sqrt{a} - \sqrt{b}\right)^{-1} - \left(\dfrac{\sqrt{a} + \sqrt{b}}{\sqrt{b}}\right)^{-1}\right]$

$\left(a + a^{\frac{1}{2}} b^{\frac{1}{2}}\right) \left(a + b\right)^{-1} \left[\sqrt{a} \left(\sqrt{a} - \sqrt{b}\right)^{-1} - \left(\dfrac{\sqrt{a} + \sqrt{b}}{\sqrt{b}}\right)^{-1}\right]$

$= \left(a + \sqrt{a} \sqrt{b}\right) \times \dfrac{1}{a + b} \times \left[\dfrac{\sqrt{a}}{\sqrt{a} - \sqrt{b}} - \dfrac{\sqrt{b}}{\sqrt{a} + \sqrt{b}}\right]$

$= \left(\dfrac{a + \sqrt{ab}}{a + b}\right) \left[\dfrac{\sqrt{a} \left(\sqrt{a} + \sqrt{b}\right) - \sqrt{b} \left(\sqrt{a} - \sqrt{b}\right)}{\left(\sqrt{a} + \sqrt{b}\right) \left(\sqrt{a} - \sqrt{b}\right)}\right]$

$= \left(\dfrac{a + \sqrt{ab}}{a + b}\right) \left[\dfrac{a + \sqrt{ab} - \sqrt{ab} + b}{a - b}\right]$

$= \dfrac{a + \sqrt{ab}}{a - b}$

$= \dfrac{\sqrt{a} \left(\sqrt{a} + \sqrt{b}\right)}{a - b}$

$= \dfrac{\sqrt{a} \left(\sqrt{a} + \sqrt{b}\right) \left(\sqrt{a} - \sqrt{b}\right)}{\left(a - b\right) \left(\sqrt{a} - \sqrt{b}\right)}$

$= \dfrac{\sqrt{a} \left(a - b\right)}{\left(a - b\right) \left(\sqrt{a} - \sqrt{b}\right)}$

$= \dfrac{\sqrt{a}}{\sqrt{a} - \sqrt{b}}$

Algebra - Algebraic Expressions

Simplify: $\;$ $\left(\dfrac{\sqrt{a}}{2} - \dfrac{1}{2 \sqrt{a}}\right)^2 \left(\dfrac{\sqrt{a} - 1}{\sqrt{a} + 1} - \dfrac{\sqrt{a} + 1}{\sqrt{a} - 1}\right)$

$\left(\dfrac{\sqrt{a}}{2} - \dfrac{1}{2 \sqrt{a}}\right)^2 \left(\dfrac{\sqrt{a} - 1}{\sqrt{a} + 1} - \dfrac{\sqrt{a} + 1}{\sqrt{a} - 1}\right)$

$= \left(\dfrac{2a - 2}{4 \sqrt{a}}\right)^2 \left[\dfrac{\left(\sqrt{a} - 1\right)^2 - \left(\sqrt{a} + 1\right)^2}{\left(\sqrt{a} + 1\right) \left(\sqrt{a} - 1\right)}\right]$

$= \left(\dfrac{a - 1}{2 \sqrt{a}}\right)^2 \left[\dfrac{a - 2 \sqrt{a} + 1 - \left(a + 2 \sqrt{a} + 1\right)}{a - 1}\right]$

$= \dfrac{a - 1}{4a} \times \left(- 4 \sqrt{a}\right)$

$= \dfrac{\left(1 - a\right) \sqrt{a}}{a}$

$= \dfrac{\left(1 - a\right) a}{a \sqrt{a}}$

$= \dfrac{1 - a}{\sqrt{a}}$

Algebra - Algebraic Expressions

Simplify: $\;$ $\dfrac{\dfrac{1}{\sqrt{a - 1}} + \sqrt{a + 1}}{\dfrac{1}{\sqrt{a + 1}} - \dfrac{1}{\sqrt{a - 1}}} : \dfrac{\sqrt{a + 1}}{\left(a - 1\right) \sqrt{a + 1} - \left(a + 1\right) \sqrt{a - 1}}$

$\dfrac{\dfrac{1}{\sqrt{a - 1}} + \sqrt{a + 1}}{\dfrac{1}{\sqrt{a + 1}} - \dfrac{1}{\sqrt{a - 1}}} : \dfrac{\sqrt{a + 1}}{\left(a - 1\right) \sqrt{a + 1} - \left(a + 1\right) \sqrt{a - 1}}$

$= \left\{\dfrac{\dfrac{1 + \sqrt{a + 1} \sqrt{a - 1}}{\sqrt{a - 1}}}{\dfrac{\sqrt{a - 1} - \sqrt{a + 1}}{\sqrt{a + 1} \sqrt{a - 1}} } \right\} \times \left\{\dfrac{\left(a - 1\right) \sqrt{a + 1} - \left(a + 1\right) \sqrt{a - 1}}{\sqrt{a + 1}} \right\}$

$= \left\{\dfrac{\left[1 + \sqrt{a + 1} \sqrt{a - 1}\right] \sqrt{a + 1}}{\sqrt{a - 1} - \sqrt{a + 1}} \right\} \times \left\{\dfrac{\left(a - 1\right) \sqrt{a + 1} - \left(a + 1\right) \sqrt{a - 1}}{\sqrt{a + 1}} \right\}$

$= \left\{\dfrac{1 + \sqrt{a + 1} \sqrt{a - 1}}{\sqrt{a - 1} - \sqrt{a + 1}} \right\} \times \left\{\sqrt{a - 1} \sqrt{a + 1} \left(\sqrt{a - 1} - \sqrt{a + 1}\right) \right\}$

$= \left\{1 + \sqrt{a + 1} \sqrt{a - 1} \right\} \times \sqrt{a - 1} \sqrt{a + 1}$

$= \sqrt{a - 1} \sqrt{a + 1} + \left(a + 1\right) \left(a - 1\right)$

$= \sqrt{a^2 - 1} + a^2 - 1$

Algebra - Algebraic Expressions

Simplify: $\;$ $\left[\dfrac{m + \sqrt{m^2 - n^2}}{m - \sqrt{m^2 - n^2}} - \dfrac{m - \sqrt{m^2 - n^2}}{m + \sqrt{m^2 - n^2}}\right] \times \dfrac{n^2}{4m \sqrt{m^2 - n^2}}$

$\left[\dfrac{m + \sqrt{m^2 - n^2}}{m - \sqrt{m^2 - n^2}} - \dfrac{m - \sqrt{m^2 - n^2}}{m + \sqrt{m^2 - n^2}}\right] \times \dfrac{n^2}{4m \sqrt{m^2 - n^2}}$

$= \left[\dfrac{\left(m + \sqrt{m^2 - n^2}\right)^2 - \left(m - \sqrt{m^2 - n^2}\right)^2}{\left(m + \sqrt{m^2 - n^2}\right) \left(m - \sqrt{m^2 - n^2}\right)}\right] \times \dfrac{n^2}{4m \sqrt{m^2 - n^2}}$

$= \left[\dfrac{\left(m + \sqrt{m^2 - n^2} + m - \sqrt{m^2 - n^2}\right) \left(m + \sqrt{m^2 - n^2} - m + \sqrt{m^2 - n^2}\right)}{m^2 - \left(m^2 - n^2\right)}\right]$
$\hspace{10cm} \times \dfrac{n^2}{4m \sqrt{m^2 - n^2}}$

$= \left[\dfrac{2m \times 2 \sqrt{m^2 - n^2}}{n^2}\right] \times \dfrac{n^2}{4m \sqrt{m^2 - n^2}}$

$= 1$

Algebra - Algebraic Expressions

Simplify: $\;$ $\left[\dfrac{1}{\left(a^{\frac{1}{2}} + b^{\frac{1}{2}}\right)^{-2}} - \left(\dfrac{\sqrt{a} - \sqrt{b}}{a^{\frac{3}{2}} - b^{\frac{3}{2}}}\right)^{-1}\right] \left(ab\right)^{\frac{-1}{2}}$

$\left[\dfrac{1}{\left(a^{\frac{1}{2}} + b^{\frac{1}{2}}\right)^{-2}} - \left(\dfrac{\sqrt{a} - \sqrt{b}}{a^{\frac{3}{2}} - b^{\frac{3}{2}}}\right)^{-1}\right] \left(ab\right)^{\frac{-1}{2}}$

$= \left[\left(a^{\frac{1}{2}} + b^{\frac{1}{2}}\right)^2 - \left(\dfrac{\left(a^{\frac{1}{2}}\right)^3 - \left(b^{\frac{1}{2}}\right)^3}{\sqrt{a} - \sqrt{b}}\right)\right] \times \dfrac{1}{\left(ab\right)^{\frac{1}{2}}}$

$= \left[\left(\sqrt{a} + \sqrt{b}\right)^2 - \dfrac{\left(\sqrt{a} - \sqrt{b}\right) \left(\left(\sqrt{a}\right)^2 + \sqrt{a} \sqrt{b} + \left(\sqrt{b}\right)^2\right)}{\sqrt{a} - \sqrt{b}} \right] \times \dfrac{1}{\sqrt{ab}}$

$= \left[a + b + 2 \sqrt{ab} - \left(a + \sqrt{ab} + b\right)\right] \times \dfrac{1}{\sqrt{ab}}$

$= \sqrt{ab} \times \dfrac{1}{\sqrt{ab}}$

$= 1$

Algebra - Algebraic Expressions

Simplify: $\;$ $2 \left(a + b\right)^{-1} \left(ab\right)^{\frac{1}{2}} \left[1 + \dfrac{1}{4} \left(\sqrt{\dfrac{a}{b}} - \sqrt{\dfrac{b}{a}}\right)^2\right]^{\frac{1}{2}}$ $\;\;$ for $a > 0, \; b > 0$

$2 \left(a + b\right)^{-1} \left(ab\right)^{\frac{1}{2}} \left[1 + \dfrac{1}{4} \left(\sqrt{\dfrac{a}{b}} - \sqrt{\dfrac{b}{a}}\right)^2\right]^{\frac{1}{2}}$

$= \dfrac{2}{a + b} \times \sqrt{ab} \times \left[1 + \dfrac{1}{4} \left(\dfrac{a}{b} - 2 \sqrt{\dfrac{a}{b}} \times \sqrt{\dfrac{b}{a}} + \dfrac{b}{a}\right) \right]^{\frac{1}{2}}$

$= \dfrac{2}{a + b} \times \sqrt{ab} \times \left[1 + \dfrac{a}{4b} - \dfrac{1}{2} + \dfrac{b}{4a}\right]^{\frac{1}{2}}$

$= \dfrac{2}{a + b} \times \sqrt{ab} \times \left[\dfrac{1}{2} + \dfrac{a}{4b} + \dfrac{b}{4a}\right]^{\frac{1}{2}}$

$= \dfrac{2}{a + b} \times \sqrt{ab} \times \left[\dfrac{2ab + a^2 + b^2}{4ab}\right]^{\frac{1}{2}}$

$= \dfrac{\left[\left(a + b\right)^2\right]^{\frac{1}{2}}}{a + b}$

$= \dfrac{a + b}{a + b}$

$= 1$ $\;\;\;$ [when $a > 0, \; b > 0 \;$ then $\; a + b > 0$]

Algebra - Algebraic Expressions

Simplify: $\;$ $\left(2^{\frac{3}{2}} + 27 y^{\frac{3}{5}}\right) : \left[\left(\dfrac{1}{2}\right)^{\frac{-1}{2}} + 3 y^{\frac{1}{5}}\right]$

$\left(2^{\frac{3}{2}} + 27 y^{\frac{3}{5}}\right) : \left[\left(\dfrac{1}{2}\right)^{\frac{-1}{2}} + 3 y^{\frac{1}{5}}\right]$

$= \dfrac{\left(2^{\frac{1}{2}}\right)^3 + \left(3 y^{\frac{1}{5}}\right)^3}{2^{\frac{1}{2}} + 3 y^{\frac{1}{5}}}$

$= \dfrac{\left(2^{\frac{1}{2}} + 3y^{\frac{1}{5}}\right) \left[\left(2^{\frac{1}{2}}\right)^2 - 2^{\frac{1}{2}} \times 3 y^{\frac{1}{5}} + \left(3y^{\frac{1}{5}}\right)^2 \right]}{2^{\frac{1}{2}} + 3 y^{\frac{1}{5}}}$

$= 2 - 3 \sqrt{2} y^{\frac{1}{5}} + 9 y^{\frac{2}{5}}$

Algebra - Algebraic Expressions

Simplify: $\;$ $\dfrac{a \sqrt{a} + b \sqrt{b}}{\left(\sqrt{a} + \sqrt{b}\right) \left(a - b\right)} + \dfrac{2 \sqrt{b}}{\sqrt{a} + \sqrt{b}} - \dfrac{\sqrt{ab}}{a - b}$

$\dfrac{a \sqrt{a} + b \sqrt{b}}{\left(\sqrt{a} + \sqrt{b}\right) \left(a - b\right)} + \dfrac{2 \sqrt{b}}{\sqrt{a} + \sqrt{b}} - \dfrac{\sqrt{ab}}{a - b}$

$= \dfrac{a \sqrt{a} + b \sqrt{b} + 2 \sqrt{b} \left(a - b\right) - \sqrt{ab} \left(\sqrt{a} + \sqrt{b}\right)}{\left(a - b\right) \left(\sqrt{a} + \sqrt{b}\right)}$

$= \dfrac{a \sqrt{a} + b \sqrt{b} + 2a \sqrt{b} - 2b \sqrt{b} - a \sqrt{b} - b \sqrt{a}}{\left(a - b\right) \left(\sqrt{a} + \sqrt{b}\right)}$

$= \dfrac{\left(a - b\right) \sqrt{a} + \left(b - a\right) \sqrt{b} + 2 \sqrt{b} \left(a - b\right)}{\left(a - b\right) \left(\sqrt{a} + \sqrt{b}\right)}$

$= \dfrac{\left(a - b\right) \left(\sqrt{a} + 2 \sqrt{b} - \sqrt{b}\right)}{\left(a - b\right) \left(\sqrt{a} + \sqrt{b}\right)}$

$= \dfrac{\sqrt{a} + \sqrt{b}}{\sqrt{a} + \sqrt{b}}$

$= 1$

Algebra - Algebraic Expressions

Simplify: $\;$ $\left[\dfrac{a^{\frac{1}{2}} + 2}{a + 2 a ^{\frac{1}{2}} + 1} - \dfrac{a^{\frac{1}{2}} - 2}{a - 1}\right] \times \left[\dfrac{a^{\frac{1}{2}} + 1}{a^{\frac{1}{2}}}\right]$

$\left[\dfrac{a^{\frac{1}{2}} + 2}{a + 2 a ^{\frac{1}{2}} + 1} - \dfrac{a^{\frac{1}{2}} - 2}{a - 1}\right] \times \left[\dfrac{a^{\frac{1}{2}} + 1}{a^{\frac{1}{2}}}\right]$

$= \left[\dfrac{\sqrt{a} + 2}{a + 2 \sqrt{a} + 1} - \dfrac{\sqrt{a} - 2}{a - 1}\right] \times \left[\dfrac{\sqrt{a} + 1}{\sqrt{a}}\right]$

$= \left[\dfrac{\left(\sqrt{a} + 2\right) \left(a - 1\right) - \left(\sqrt{a} - 2\right) \left(a + 2 \sqrt{a} + 1\right)}{\left(\sqrt{a} + 1\right)^2 \left(a - 1\right)}\right] \times \left[\dfrac{\sqrt{a} + 1}{\sqrt{a}}\right]$

$= \dfrac{a \sqrt{a} - \sqrt{a} + 2a - 2 - \left(a \sqrt{a} + 2a + \sqrt{a} - 2a - 4 \sqrt{a} - 2\right)}{\sqrt{a} \left(\sqrt{a} + 1\right) \left(a - 1\right)}$

$= \dfrac{2a + 2 \sqrt{a}}{\sqrt{a} \left(\sqrt{a} + 1\right) \left(a - 1\right)}$

$= \dfrac{2 \sqrt{a} \left(\sqrt{a} + 1\right)}{\sqrt{a} \left(\sqrt{a} + 1\right) \left(a - 1\right)}$

$= \dfrac{2}{a - 1}$

Algebra - Algebraic Expressions

Simplify: $\;$ $\left[\left(\dfrac{a \sqrt{a} + b \sqrt{b}}{\sqrt{a} + \sqrt{b}} - \sqrt{ab}\right) : \left(a - b\right)\right] + \dfrac{2 \sqrt{b}}{\sqrt{a} + \sqrt{b}}$

$\left[\left(\dfrac{a \sqrt{a} + b \sqrt{b}}{\sqrt{a} + \sqrt{b}} - \sqrt{ab}\right) : \left(a - b\right)\right] + \dfrac{2 \sqrt{b}}{\sqrt{a} + \sqrt{b}}$

$= \left[\dfrac{\sqrt{a} \left(a - b\right) + \sqrt{b} \left(b - a\right)}{\sqrt{a} + \sqrt{b}} : \left(a - b\right)\right] + \dfrac{2 \sqrt{b}}{\sqrt{a} + \sqrt{b}}$

$= \left[\dfrac{\left(\sqrt{a} - \sqrt{b}\right) \left(a - b\right)}{\sqrt{a} + \sqrt{b}} \times \dfrac{1}{a - b}\right] + \dfrac{2 \sqrt{b}}{\sqrt{a} + \sqrt{b}}$

$= \dfrac{\sqrt{a} - \sqrt{b}}{\sqrt{a} + \sqrt{b}} + \dfrac{2 \sqrt{b}}{\sqrt{a} + \sqrt{b}}$

$= \dfrac{\sqrt{a} + \sqrt{b}}{\sqrt{a} + \sqrt{b}}$

$= 1$

Algebra - Algebraic Expressions

Simplify: $\;$ $x^{\frac{1}{2}} + x^{\frac{-1}{2}} + \dfrac{\left(1 - x\right) \left(1 - x^{\frac{-1}{2}}\right)}{1 + \sqrt{x}}$

$x^{\frac{1}{2}} + x^{\frac{-1}{2}} + \dfrac{\left(1 - x\right) \left(1 - x^{\frac{-1}{2}}\right)}{1 + \sqrt{x}}$

$= \sqrt{x} + \dfrac{1}{\sqrt{x}} + \dfrac{\left(1 - x\right) \left(1 - \dfrac{1}{\sqrt{x}}\right)}{1 + \sqrt{x}}$

$= \dfrac{x + 1}{\sqrt{x}} + \dfrac{1 - \dfrac{1}{\sqrt{x}} - x + \dfrac{x}{\sqrt{x}}}{1 + \sqrt{x}}$

$= \dfrac{x + 1}{\sqrt{x}} + \dfrac{\sqrt{x} - 1 - x \sqrt{x} + x}{\sqrt{x} \left(1 + \sqrt{x}\right)}$

$= \dfrac{\left(x + 1\right) \left(1 + \sqrt{x}\right) + \sqrt{x} - 1 - x \sqrt{x} + x}{\sqrt{x} \left(1 + \sqrt{x}\right)}$

$= \dfrac{x + x \sqrt{x} + 1 + \sqrt{x} + \sqrt{x} - 1 - x \sqrt{x} + x}{\sqrt{x} \left(1 + \sqrt{x}\right)}$

$= \dfrac{2x + 2 \sqrt{x}}{\sqrt{x} + x}$

$= \dfrac{2 \left(\sqrt{x} + x\right)}{\sqrt{x} + x}$

$= 2$

Algebra - Algebraic Expressions

Simplify: $\;$ $\dfrac{a - a^{-2}}{a^{\frac{1}{2}} - a^{\frac{-1}{2}}} - \dfrac{2}{a^{\frac{3}{2}}} - \dfrac{1 - a^{-2}}{a^{\frac{1}{2}} + a^{\frac{-1}{2}}}$

$\dfrac{a - a^{-2}}{a^{\frac{1}{2}} - a^{\frac{-1}{2}}} - \dfrac{2}{a^{\frac{3}{2}}} - \dfrac{1 - a^{-2}}{a^{\frac{1}{2}} + a^{\frac{-1}{2}}}$

$= \dfrac{a - \dfrac{1}{a^2}}{\sqrt{a} - \dfrac{1}{\sqrt{a}}} - \dfrac{2}{\left(a^3\right)^{\frac{1}{2}}} - \dfrac{1 - \dfrac{1}{a^2}}{\sqrt{a} + \dfrac{1}{\sqrt{a}}}$

$= \dfrac{\dfrac{a^3 - 1}{a^2}}{\dfrac{a - 1}{\sqrt{a}}} - \dfrac{2}{\sqrt{a^3}} - \dfrac{\dfrac{a^2 - 1}{a^2}}{\dfrac{a + 1}{\sqrt{a}}}$

$= \dfrac{\left(a - 1\right) \left(a^2 + a + 1\right) \sqrt{a}}{a^2 \left(a - 1\right)} - \dfrac{2}{a \sqrt{a}} - \dfrac{\left(a + 1\right) \left(a - 1\right) \sqrt{a}}{a^2 \left(a + 1\right)}$

$= \dfrac{\left(a^2 + a + 1\right) \sqrt{a}}{a^2} - \dfrac{2}{a \sqrt{a}} - \dfrac{\left(a - 1\right) \sqrt{a}}{a^2}$

$= \dfrac{a^2 \sqrt{a} + a \sqrt{a} + \sqrt{a} - 2 \sqrt{a} - a \sqrt{a} + \sqrt{a}}{a^2}$

$= \dfrac{a^2 \sqrt{a}}{a^2}$

$= \sqrt{a}$

Algebra - Algebraic Expressions

Simplify: $\;$ $\dfrac{x^{\frac{1}{2}} + x^{\frac{-1}{2}}}{1 - x} + \dfrac{1 - x^{\frac{-1}{2}}}{1 + \sqrt{x}}$

$\dfrac{x^{\frac{1}{2}} + x^{\frac{-1}{2}}}{1 - x} + \dfrac{1 - x^{\frac{-1}{2}}}{1 + \sqrt{x}}$

$= \dfrac{\sqrt{x} + \dfrac{1}{\sqrt{x}}}{1 - x} + \dfrac{1 - \dfrac{1}{\sqrt{x}}}{1 + \sqrt{x}}$

$= \dfrac{x + 1}{\sqrt{x} \left(1 - x\right)} + \dfrac{\sqrt{x} - 1}{\sqrt{x} \left(1 + \sqrt{x}\right)}$

$= \dfrac{x + 1}{\sqrt{x} - x \sqrt{x}} + \dfrac{\sqrt{x} - 1}{\sqrt{x} + x}$

$= \dfrac{\left(x + 1\right) \left(\sqrt{x} + x\right) + \left(\sqrt{x} - 1\right) \left(\sqrt{x} - x \sqrt{x}\right)}{\left(\sqrt{x} - x \sqrt{x}\right) \left(\sqrt{x} + x\right)}$

$= \dfrac{x \sqrt{x} + x^2 + \sqrt{x} + x + x - x^2 - \sqrt{x} + x \sqrt{x}}{x + x \sqrt{x} - x^2 - x^2 \sqrt{x}}$

$= \dfrac{2x + 2x \sqrt{x}}{x \left(1 + \sqrt{x}\right) - x^2 \left(1 + \sqrt{x}\right)}$

$= \dfrac{2x \left(1 + \sqrt{x}\right)}{\left(x - x^2\right) \left(1 + \sqrt{x}\right)}$

$= \dfrac{2x}{x \left(1 - x\right)}$

$= \dfrac{2}{1 - x}$

Algebra - Algebraic Expressions

Simplify: $\;$ $\left(\dfrac{1}{2 + 2 \sqrt{a}} + \dfrac{1}{2 - 2 \sqrt{a}} - \dfrac{a^2 + 1}{1 - a^2}\right) \left(1 + \dfrac{1}{a}\right)$

$\left(\dfrac{1}{2 + 2 \sqrt{a}} + \dfrac{1}{2 - 2 \sqrt{a}} - \dfrac{a^2 + 1}{1 - a^2}\right) \left(1 + \dfrac{1}{a}\right)$

$= \left(\dfrac{2 - 2 \sqrt{a} + 2 + 2 \sqrt{a}}{\left(2 + 2 \sqrt{a}\right) \left(2 - 2 \sqrt{a}\right)} - \dfrac{a^2 + 1}{1 - a^2}\right) \left(\dfrac{a + 1}{a}\right)$

$= \left(\dfrac{4}{4 - 4a} - \dfrac{a^2 + 1}{\left(1 + a\right) \left(1 - a\right)}\right) \left(\dfrac{a + 1}{a}\right)$

$= \left(\dfrac{1 + a - a^2 - 1}{\left(1 + a\right) \left(1 - a\right)}\right) \left(\dfrac{a + 1}{a}\right)$

$= \dfrac{a \left(1 - a\right)}{1 - a} \times \dfrac{1}{a}$

$= 1$

Algebra - Algebraic Expressions

Simplify: $\;$ $\dfrac{\sqrt[3]{a^5 b^{\frac{1}{2}} \sqrt[4]{a^{-1}}}}{\left(a^2 \sqrt[5]{ab^3}\right)^2}$

$\dfrac{\sqrt[3]{a^5 b^{\frac{1}{2}} \sqrt[4]{a^{-1}}}}{\left(a^2 \sqrt[5]{ab^3}\right)^2}$

$= \dfrac{\sqrt[3]{a^5 \times b^{\frac{1}{2}} \times a^{\frac{-1}{4}}}}{\left(a^2 \times a^{\frac{1}{5}} \times b^{\frac{3}{5}}\right)^2}$

$= \dfrac{\left(a^{\frac{19}{4}} \times b^{\frac{1}{2}}\right)^{\frac{1}{3}}}{\left(a^{\frac{11}{5}} \times b^{\frac{3}{5}}\right)^2}$

$= \dfrac{a^{\frac{19}{12}} \times b^{\frac{1}{6}}}{a^{\frac{22}{5}} \times b^{\frac{6}{5}}}$

$= a^{\frac{19}{12} - \frac{22}{5}} \times b^{\frac{1}{6} - \frac{6}{5}}$

$= a^{\frac{-169}{60}} \times b^{\frac{-31}{30}}$

Algebra - Algebraic Expressions

Simplify: $\;$ $\dfrac{2b + a - \dfrac{4a^2 - b^2}{a}}{b^3 + 2ab^2 - 3a^2 b} \times \dfrac{a^3 b - 2a^2 b^2 + ab^3}{a^2 - b^2}$

$\dfrac{2b + a - \dfrac{4a^2 - b^2}{a}}{b^3 + 2ab^2 - 3a^2 b} \times \dfrac{a^3 b - 2a^2 b^2 + ab^3}{a^2 - b^2}$

$= \dfrac{2ab + a^2 - 4a^2 + b^2}{ab \left(b^2 + 2ab - 3a^2\right)} \times \dfrac{ab \left(a^2 - 2ab + b^2\right)}{\left(a + b\right) \left(a - b\right)}$

$= \dfrac{2ab - 3a^2 + b^2}{b^2 + 2ab - 3a^2} \times \dfrac{\left(a - b\right)^2}{\left(a + b\right) \left(a - b\right)}$

$= \dfrac{a - b}{a + b}$

Algebra - Algebraic Expressions

Simplify: $\;$ $\left(\dfrac{\sqrt{2}}{\left(1 - x^2\right)^{-1}} + \dfrac{2^{3/2}}{x^{-2}}\right) : \left(\dfrac{x^{-2}}{1 + x^{-2}}\right)^{-1}$

$\left(\dfrac{\sqrt{2}}{\left(1 - x^2\right)^{-1}} + \dfrac{2^{3/2}}{x^{-2}}\right) : \left(\dfrac{x^{-2}}{1 + x^{-2}}\right)^{-1}$

$= \left(\sqrt{2} \left(1 - x^2\right) + 2 \sqrt{2} x^2\right) : \left(\dfrac{1 + x^{-2}}{x^{-2}}\right)$

$= \left(\sqrt{2} - \sqrt{2} x^2 + 2 \sqrt{2} x^2\right) : \left(\dfrac{1 + \dfrac{1}{x^2}}{\dfrac{1}{x^2}}\right)$

$= \left(\sqrt{2} + \sqrt{2} x^2\right) : \left(x^2 + 1\right)$

$= \dfrac{\sqrt{2} \left(1 + x^2\right)}{1 + x^2}$

$= \sqrt{2}$

Algebra - Algebraic Expressions

Simplify: $\;$ $a - \left\{\left[\dfrac{\left(16 - a\right)a}{a^2 - 4} + \dfrac{3 + 2a}{2 - a} + \dfrac{3a - 2}{a + 2} \right] : \dfrac{a - 1}{a \left(a^2 + 4a + 4\right)} \right\}$

$a - \left\{\left[\dfrac{\left(16 - a\right)a}{a^2 - 4} + \dfrac{3 + 2a}{2 - a} + \dfrac{3a - 2}{a + 2} \right] : \dfrac{a - 1}{a \left(a^2 + 4a + 4\right)} \right\}$

$= a - \left\{\left[\dfrac{\left(16 - a\right) a}{a^2 - 4} + \dfrac{\left(3 + 2a\right) \left(2 + a\right) + \left(3a - 2\right) \left(2 - a\right)}{\left(2 + a\right) \left(2 - a\right)} \right] : \dfrac{a - 1}{a \left(a + 2\right)^2} \right\}$

$= a - \left\{\left[\dfrac{\left(16 - a\right) a}{a^2 - 4} + \dfrac{6 + 3a + 4a + 2a^2 + 6a - 3a^2 - 4 + 2a}{4 - a^2}\right] : \dfrac{a - 1}{a \left(a + 2\right)^2} \right\}$

$= a - \left\{\left[\dfrac{\left(16 - a\right) a}{a^2 - 4} + \dfrac{2 + 15a - a^2}{4 - a^2}\right] : \dfrac{a - 1}{a \left(a + 2\right)^2} \right\}$

$= a - \left\{ \dfrac{16a - a^2 - 2 - 15a + a^2}{a^2 - 4} : \dfrac{a - 1}{a \left(a + 2\right)^2}\right\}$

$= a - \left\{\dfrac{a - 2}{a^2 - 4} : \dfrac{a - 1}{a \left(a + 2\right)^2} \right\}$

$= a - \left\{\dfrac{a - 2}{\left(a + 2\right) \left(a - 2\right)} : \dfrac{a - 1}{a \left(a + 2\right)^2} \right\}$

$= a - \left\{\dfrac{1}{a + 2} \times \dfrac{a \left(a + 2\right)^2}{a - 1} \right\}$

$= a - \dfrac{a \left(a + 2\right)}{a - 1}$

$= \dfrac{a^2 - a - a^2 - 2a}{a - 1}$

$= \dfrac{-3a}{a - 1}$

Algebra - Algebraic Expressions

Simplify: $\;$ $\left[\left(\dfrac{y}{y - x}\right)^{-2} - \dfrac{\left(x + y\right)^2 - 4xy}{x^2 - xy}\right] \dfrac{x^4}{x^2 y^2 - y^4}$

$\left[\left(\dfrac{y}{y - x}\right)^{-2} - \dfrac{\left(x + y\right)^2 - 4xy}{x^2 - xy}\right] \dfrac{x^4}{x^2 y^2 - y^4}$

$= \left[\left(\dfrac{y - x}{y}\right)^2 - \dfrac{x^2 + y^2 + 2xy - 4xy}{x \left(x - y\right)}\right] \dfrac{x^4}{y^2 \left(x^2 - y^2\right)}$

$= \left[\dfrac{\left(y - x\right)^2}{y^2} - \dfrac{x^2 + y^2 - 2xy}{x \left(x - y\right)}\right] \dfrac{x^4}{y^2 \left(x + y\right) \left(x - y\right)}$

$= \left[\dfrac{\left(y - x\right)^2}{y^2} - \dfrac{\left(y - x\right)^2}{x \left(x - y\right)}\right] \dfrac{x^4}{y^2 \left(x + y\right) \left(x - y\right)}$

$= \left(y - x\right)^2 \left[\dfrac{1}{y^2} - \dfrac{1}{x \left(x - y\right)}\right] \dfrac{x^4}{y^2 \left(x + y\right) \left(x - y\right)}$

$= \dfrac{\left(y - x\right)^2 \left(x^2 - xy - y^2\right) x^4}{y^2 x \left(x - y\right) y^2 \left(x + y\right) \left(x - y\right)}$

$= \dfrac{\left(y - x\right)^2 \left(x^2 - xy - y^2\right) x^3}{y^4 \left(x + y\right) \left(x - y\right)^2}$

$= \dfrac{x^3 \left(x^2 - xy - y^2\right)}{y^4 \left(x + y\right)}$

Algebra - Algebraic Expressions

Simplify: $\;$ $\left[\dfrac{x^3 + y^3}{x + y} : \left(x^2 - y^2\right)\right]+ \left[\dfrac{2y}{x + y} - \dfrac{xy}{x^2 - y^2}\right]$

$\left[\dfrac{x^3 + y^3}{x + y} : \left(x^2 - y^2\right)\right]+ \left[\dfrac{2y}{x + y} - \dfrac{xy}{x^2 - y^2}\right]$

$= \left[\dfrac{\left(x + y\right) \left(x^2 - xy + y^2\right)}{x + y} : \left(x^2 - y^2\right)\right] + \left[\dfrac{2y}{x + y} - \dfrac{xy}{\left(x + y\right) \left(x - y\right)}\right]$

$= \dfrac{x^2 - xy + y^2}{x^2 - y^2} + \dfrac{1}{x + y} \left[2y - \dfrac{xy}{x - y}\right]$

$= \dfrac{x^2 - xy + y^2}{x^2 - y^2} + \dfrac{2xy - 2y^2 - xy}{\left(x + y\right) \left(x - y\right)}$

$= \dfrac{x^2 - xy + y^2}{x^2 - y^2} + \dfrac{xy - 2y^2}{x^2 - y^2}$

$= \dfrac{x^2 - xy + y^2 + xy - 2y^2}{x^2 - y^2}$

$= \dfrac{x^2 - y^2}{x^2 - y^2}$

$= 1$

Algebra - Algebraic Expressions

Simplify: $\;$ $\left(\dfrac{\dfrac{1}{a} + \dfrac{1}{b + c}}{\dfrac{1}{a} - \dfrac{1}{b + c}}\right) \left(1 + \dfrac{b^2 + c^2 - a^2}{2bc}\right) \left(a + b + c\right)^{-2}$

$\left(\dfrac{\dfrac{1}{a} + \dfrac{1}{b + c}}{\dfrac{1}{a} - \dfrac{1}{b + c}}\right) \left(1 + \dfrac{b^2 + c^2 - a^2}{2bc}\right) \left(a + b + c\right)^{-2}$

$= \left(\dfrac{a + b + c}{b + c - a}\right) \left(\dfrac{2bc + b^2 + c^2 - a^2}{2bc}\right) \left(\dfrac{1}{a + b + c}\right)^2$

$= \dfrac{\left(b + c\right)^2 - a^2}{2bc \left(b + c + a\right) \left(b + c - a\right)}$

$= \dfrac{\left(b + c\right)^2 - a^2}{2bc \left[\left(b + c\right)^2 - a^2\right]}$

$= \dfrac{1}{2bc}$

Algebra - Algebraic Expressions

Simplify: $\;$ $\left[\dfrac{1}{\left(m + n\right)^2} \left(\dfrac{1}{m^2} + \dfrac{1}{n^2}\right) + \dfrac{2}{\left(m + n\right)^3} \left(\dfrac{1}{m} + \dfrac{1}{n}\right)\right] m^2 n^2$

$\left[\dfrac{1}{\left(m + n\right)^2} \left(\dfrac{1}{m^2} + \dfrac{1}{n^2}\right) + \dfrac{2}{\left(m + n\right)^3} \left(\dfrac{1}{m} + \dfrac{1}{n}\right)\right] m^2 n^2$

$= \left[\dfrac{1}{\left(m + n\right)^2} \left(\dfrac{m^2 + n^2}{m^2 n^2}\right) + \dfrac{2}{\left(m + n\right)^3} \left(\dfrac{m + n}{mn}\right)\right] m^2 n^2$

$= \left[\dfrac{1}{\left(m + n\right)^2} \left(\dfrac{m^2 + n^2}{m^2 n^2}\right) + \dfrac{2}{\left(m + n\right)^2} \times \dfrac{1}{mn}\right] m^2 n^2$

$= \dfrac{m^2 + n^2}{\left(m + n\right)^2} + \dfrac{2 mn}{\left(m + n\right)^2}$

$= \dfrac{m^2 + n^2 + 2mn}{\left(m + n\right)^2}$

$= \dfrac{\left(m + n\right)^2}{\left(m + n\right)^2}$

$= 1$

Algebra - Algebraic Expressions

Simplify: $\;$ $\left[\dfrac{a + 3b}{\left(a - b\right)^2} + \dfrac{a - 3b}{a^2 - b^2}\right] : \dfrac{a^2 + 3b^2}{\left(a - b\right)^2}$

$\left[\dfrac{a + 3b}{\left(a - b\right)^2} + \dfrac{a - 3b}{a^2 - b^2}\right] : \dfrac{a^2 + 3b^2}{\left(a - b\right)^2}$

$= \dfrac{\left(a + 3b\right) \left(a^2 - b^2\right) + \left(a - 3b\right) \left(a - b\right)^2}{\left(a - b\right)^2 \left(a^2 - b^2\right)} \times \dfrac{\left(a - b\right)^2}{a^2 + 3b^2}$

$= \dfrac{\left(a + 3b\right) \left(a + b\right) \left(a - b\right) + \left(a - 3b\right) \left(a - b\right)^2}{\left(a^2 - b^2\right) \left(a^2 + 3b^2\right)}$

$= \dfrac{\left(a - b\right) \left[\left(a + 3b\right) \left(a + b\right) + \left(a - 3b\right) \left(a - b\right)\right]}{\left(a + b\right) \left(a - b\right) \left(a^2 + 3b^2\right)}$

$= \dfrac{a^2 + ab + 3ab + 3b^2 + a^2 - ab - 3ab + 3b^2}{\left(a + b\right) \left(a^2 + 3b^2\right)}$

$= \dfrac{2a^2 + 6b^2}{\left(a + b\right) \left(a^2 + 3b^2\right)}$

$= \dfrac{2 \left(a^2 + 3b^2\right)}{\left(a + b\right) \left(a^2 + 3b^2\right)}$

$= \dfrac{2}{a + b}$

Algebra - Algebraic Expressions

Simplify: $\;$ $\left(a + \dfrac{ab}{a - b}\right) \left(\dfrac{ab}{a + b} - a\right) : \dfrac{a^2 + b^2}{a^2 - b^2}$

$\left(a + \dfrac{ab}{a - b}\right) \left(\dfrac{ab}{a + b} - a\right) : \dfrac{a^2 + b^2}{a^2 - b^2}$

$= \left(\dfrac{a^2 - ab + ab}{a - b}\right) \left(\dfrac{ab - a^2 - ab}{a + b}\right) : \dfrac{a^2 + b^2}{a^2 - b^2}$

$= \dfrac{-a^4}{a^2 - b^2} : \dfrac{a^2 + b^2}{a^2 - b^2}$

$= \dfrac{-a^4}{a^2 + b^2}$

Algebra - Algebraic Expressions

Simplify: $\;$ $\dfrac{a^2 - b^2}{a - b} - \dfrac{a^3 - b^3}{a^2 - b^2}$

$\dfrac{a^2 - b^2}{a - b} - \dfrac{a^3 - b^3}{a^2 - b^2}$

$= \dfrac{\left(a + b\right) \left(a - b\right)}{a - b} - \dfrac{\left(a - b\right) \left(a^2 + ab + b^2\right)}{\left(a + b\right) \left(a - b\right)}$

$= a + b - \dfrac{a^2 + ab + b^2}{a + b}$

$= \dfrac{\left(a + b\right)^2 - \left(a^2 + ab + b^2\right)}{a + b}$

$= \dfrac{a^2 + 2ab + b^2 - a^2 - ab - b^2}{a + b}$

$= \dfrac{ab}{a + b}$

Algebra - Algebraic Expressions

Calculate $\;$ $x^3 + 3x - 14$ $\;$ for $\;$ $x = \sqrt[3]{7 + 5 \sqrt{2}} - \dfrac{1}{\sqrt[3]{7 + 5 \sqrt{2}}}$

$x^3 + 3x - 14$

$= \left(\sqrt[3]{7 + 5 \sqrt{2}} - \dfrac{1}{\sqrt[3]{7 + 5 \sqrt{2}}}\right)^3 + 3 \left(\sqrt[3]{7 + 5 \sqrt{2}} - \dfrac{1}{\sqrt[3]{7 + 5 \sqrt{2}}}\right) - 14$

$= \left(\sqrt[3]{7 + 5 \sqrt{2}}\right)^3 - 3 \times \left(\sqrt[3]{7 + 5 \sqrt{2}}\right)^2 \times \dfrac{1}{\sqrt[3]{7 + 5 \sqrt{2}}}$
$\hspace{1cm}$ $+ 3 \times \sqrt[3]{7 + 5 \sqrt{2}} \times \dfrac{1}{\left(\sqrt[3]{7 + 5 \sqrt{2}}\right)^2} - \left(\dfrac{1}{\sqrt[3]{7 + 5\sqrt{2}}}\right)^3$
$\hspace{3.5cm} + 3 \times \sqrt[3]{7 + 5 \sqrt{2}} - \dfrac{3}{\sqrt[3]{7 + 5 \sqrt{2}}} - 14$

$= 7 + 5 \sqrt{2} - 3 \sqrt[3]{7 + 5 \sqrt{2}} + \dfrac{3}{\sqrt[3]{7 + 5 \sqrt{2}}} - \dfrac{1}{7 + 5 \sqrt{2}}$
$\hspace{3.5cm} + 3 \sqrt[3]{7 + 5 \sqrt{2}} - \dfrac{3}{\sqrt[3]{7 + 5 \sqrt{2}}} - 14$

$= 7 + 5 \sqrt{2} - \dfrac{1}{7 + 5 \sqrt{2}} - 14$

$= 7 + 5 \sqrt{2} - \dfrac{\left(7 - 5 \sqrt{2}\right)}{\left(7 + 5 \sqrt{2}\right) \left(7 - 5 \sqrt{2}\right)} - 14$

$= 7 + 5 \sqrt{2} - \dfrac{\left(7 - 5 \sqrt{2}\right)}{49 - 50} - 14$

$= 7 + 5 \sqrt{2} + 7 - 5 \sqrt{2} - 14$

$= 0$

Algebra - Algebraic Expressions

Calculate $\;$ $\left[\left(128^{\frac{3}{7}} \times 27^{\frac{1}{3}} \times 10^{- \log 48}\right)^{\frac{-1}{2}} + \left(\cot \dfrac{2 \pi}{3}\right)^{-1}\right]^2 + 2 \times 6^{\frac{1}{2}}$

$\left[\left(128^{\frac{3}{7}} \times 27^{\frac{1}{3}} \times 10^{- \log 48}\right)^{\frac{-1}{2}} + \left(\cot \dfrac{2 \pi}{3}\right)^{-1}\right]^2 + 2 \times 6^{\frac{1}{2}}$

$= \left[\left(2^{7 \times \frac{3}{7}} \times 3^{3 \times \frac{1}{3}} \times 10^{\log \left(\frac{1}{48}\right)}\right)^{\frac{-1}{2}} + \dfrac{1}{\cot \dfrac{2 \pi}{3}}\right]^2 + 2 \sqrt{6}$
$\hspace{3cm}$ $\left[\text{Note: } \log n^m = m \log n\right]$

$= \left[\left(2^3 \times 3 \times \dfrac{1}{48}\right)^{\frac{-1}{2}} + \tan \dfrac{2 \pi}{3}\right]^2 + 2 \sqrt{6}$ $\;\;\;$ $\left[\text{Note: } a^{\log_a x} = x\right]$

$= \left[\left(\dfrac{1}{2}\right)^{\frac{-1}{2}} - \sqrt{3}\right]^2 + 2 \sqrt{6}$

$= \left[\left(\dfrac{1}{2^{\frac{1}{2}}}\right)^{-1} - \sqrt{3}\right]^2 + 2 \sqrt{6}$

$= \left[2^{\frac{1}{2}} - \sqrt{3}\right]^2 + 2 \sqrt{6}$

$= \left[\sqrt{2} - \sqrt{3}\right]^2 + 2 \sqrt{6}$

$= 2 + 3 - 2 \sqrt{6} + 2 \sqrt{6}$

$= 5$

Algebra - Algebraic Expressions

Calculate $\;$ $\left(\dfrac{x - 1}{x^{\frac{3}{4}} + x^{\frac{1}{2}}}\right) \cdot \left(\dfrac{x^{\frac{1}{2}} + x^{\frac{1}{4}}}{x^{\frac{1}{2}} + 1}\right) \cdot x^{\frac{1}{4}} + 1$ $\;\;\;$ for $\;$ $x = 16$

$\left(\dfrac{x - 1}{x^{\frac{3}{4}} + x^{\frac{1}{2}}}\right) \cdot \left(\dfrac{x^{\frac{1}{2}} + x^{\frac{1}{4}}}{x^{\frac{1}{2}} + 1}\right) \cdot x^{\frac{1}{4}} + 1$

$= \left(\dfrac{x - 1}{x^{\frac{1}{2}} \cdot x^{\frac{1}{4}} + x^{\frac{1}{2}}}\right) \cdot \left(\dfrac{x^{\frac{1}{2}} + x^{\frac{1}{2}} \cdot x^{\frac{-1}{4}}}{x^{\frac{1}{2}} + 1}\right) \cdot x^{\frac{1}{4}} + 1$

$= \left[\dfrac{x - 1}{x^{\frac{1}{2}} \left(x^{\frac{1}{4}} + 1\right)}\right] \cdot \left[\dfrac{x^{\frac{1}{2}} \left(1 + x^{\frac{-1}{4}}\right)}{x^{\frac{1}{2}} + 1}\right] \cdot x^{\frac{1}{4}} + 1$

$= \dfrac{\left(x - 1\right) \cdot \left(1 + \dfrac{1}{x^{\frac{1}{4}}}\right) \cdot x^{\frac{1}{4}}}{\left(x^{\frac{1}{4}} + 1\right) \cdot \left(x^{\frac{1}{2}} + 1\right)} + 1$

$= \dfrac{\left(x - 1\right) \cdot \left(x^{\frac{1}{4}} + 1\right) \cdot x^{\frac{1}{4}}}{x^{\frac{1}{4}} \left(x^{\frac{1}{4}} + 1\right) \left(x^{\frac{1}{2}} + 1\right)} + 1$

$= \dfrac{x - 1}{x^{\frac{1}{2}} + 1} + 1$

$= \dfrac{\left(x - 1\right) \left(x^{\frac{1}{2}} - 1\right)}{\left(x^{\frac{1}{2}} + 1\right) \left(x^{\frac{1}{2}} - 1\right)} + 1$

$= \dfrac{\left(x - 1\right) \left(x^{\frac{1}{2}} - 1\right)}{x - 1} + 1$

$= x^{\frac{1}{2}} - 1 + 1 = x^{\frac{1}{2}}$

$= \left(16\right)^{\frac{1}{2}}$ $\;\;\;$ when $\;$ $x = 16$

$= 4$

Algebra - Algebraic Expressions

Calculate $\;$ $\left(x^{\frac{1}{3}} + y^{\frac{1}{3}}\right) \left(x^{\frac{2}{3}} - x^{\frac{1}{3}} y^{\frac{1}{3}} + y^{\frac{2}{3}}\right)$ $\;\;\;$ for $\;$ $x = 4 \dfrac{5}{7}$, $\;$ $y = 5 \dfrac{2}{7}$

Let $\;$ $x^{\frac{1}{3}} = p$ $\;\;\; \cdots \; (1a)$, $\;$ $y^{\frac{1}{3}} = q$ $\;\;\; \cdots \; (1b)$

Then,

$\begin{aligned} \left(x^{\frac{1}{3}} + y^{\frac{1}{3}}\right) \left(x^{\frac{2}{3}} - x^{\frac{1}{3}} y^{\frac{1}{3}} + y^{\frac{2}{3}}\right) & = \left(p + q\right) \left(p^2 - pq + q^2\right) \\\\ & = p^3 + q^3 \\\\ & = \left(x^{\frac{1}{3}}\right)^3 + \left(y^{\frac{1}{3}}\right)^3 \;\; \left[\text{by equations (1a) and (1b)}\right] \\\\ & = x + y \\\\ & = \dfrac{33}{7} + \dfrac{37}{7} \;\;\; \text{when } x = 4 \dfrac{5}{7} = \dfrac{33}{7}, \; y = 5 \dfrac{2}{7} = \dfrac{37}{7} \\\\ & = \dfrac{70}{7} = 10 \end{aligned}$

Algebra - Algebraic Expressions

Calculate $\;$ $\log_{\frac{1}{2}} \left[\log_3 \cos \left(\dfrac{\pi}{6}\right) - \log_3 \sin \left(\dfrac{\pi}{6}\right)\right]$

$\begin{aligned} \log_{\frac{1}{2}} \left[\log_3 \cos \left(\dfrac{\pi}{6}\right) - \log_3 \sin \left(\dfrac{\pi}{6}\right)\right] & = \log_{\frac{1}{2}} \left[\log_3 \left(\dfrac{\cos \left(\dfrac{\pi}{6}\right)}{\sin \left(\dfrac{\pi}{6}\right)}\right)\right] \\\\ & \hspace{2cm} \left\{\text{Note: } \log_a \left(\dfrac{x}{y}\right) = \log_a x - \log_a y \right\} \\\\ & = \log_{\frac{1}{2}} \left[\log_3 \cot \left(\dfrac{\pi}{6}\right)\right] \\\\ & = \log_{\frac{1}{2}} \left[\log_3 \sqrt{3}\right] \\\\ & = \log_{\frac{1}{2}} \left[\log_3 3^{\frac{1}{2}}\right] \\\\ & = \log_{\frac{1}{2}} \left[\dfrac{1}{2} \log_3 3\right] \\\\ & = \log_{\frac{1}{2}} \dfrac{1}{2} \;\; \left\{\text{Note: } \log_a a = 1 \right\} \\\\ & = 1 \end{aligned}$

Algebra - Algebraic Expressions

Calculate $\;$ $\cos \left[\dfrac{\pi}{10} \left(\log_3 \dfrac{1}{9} + \log_{\frac{1}{9}} 3\right)\right]$

$\begin{aligned} \cos \left[\dfrac{\pi}{10} \left(\log_3 \dfrac{1}{9} + \log_{\frac{1}{9}} 3\right)\right] & = \cos \left[\dfrac{\pi}{10} \left(\log_3 3^{-2} + \dfrac{\log_3 3}{\log_3 \dfrac{1}{9}}\right)\right] \\\\ & \hspace{2cm} \left\{\text{Note: }\log_n m = \dfrac{\log_a m}{\log_a n}\right\} \\\\ & = \cos\left[\dfrac{\pi}{10} \left(\log_3 3^{-2} + \dfrac{\log_3 3}{\log_3 3^{-2}}\right)\right] \\\\ & = \cos\left[\dfrac{\pi}{10} \left(-2 \log_3 3 + \dfrac{\log_3 3}{-2 \log_3 3}\right)\right] \\\\ & \hspace{2cm} \left\{\text{Note: } \log_a m^n = n \log_a m \right\} \\\\ & = \cos \left[\dfrac{\pi}{10} \left(-2 + \dfrac{1}{-2}\right)\right] \;\; \left\{\text{Note: } \log_a a = 1 \right\} \\\\ & = \cos \left[\dfrac{\pi}{10} \times \left(\dfrac{-5}{2}\right)\right] \\\\ & = \cos \left[\dfrac{- \pi}{4}\right] \\\\ & = \cos \left[\dfrac{\pi}{4}\right] \\\\ & = \dfrac{1}{\sqrt{2}} \end{aligned}$

Coordinate Geometry - Parabola

If the normal at $P \left(18, 12\right)$ to the parabola $y^2 = 8x$ cuts it again at $Q$, show that $\;$ $9PQ = 80 \sqrt{10}$

Equation of parabola: $\;$ $y^2 = 8x$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the standard equation of parabola $\;$ $y^2 = 4ax$ $\;$ gives

$4a = 8$ $\implies$ $a = 2$

Equation of normal to the parabola $\;$ $y^2 = 4ax$ $\;$ at $\;$ $P \left(x_1, y_1\right)$ $\;$ is

$y - y_1 = \left(\dfrac{-y_1}{2a}\right) \left(x - x_1\right)$

Given point: $\;$ $P \left(18, 12\right)$

$\therefore \;$ Equation of normal at $P \left(18, 12\right)$ to parabola $(1)$ is

$y - 12 = \left(\dfrac{-12}{2 \times 2}\right) \left(x - 18\right)$

i.e. $\;$ $y - 12 = -3x + 54$

i.e. $\;$ $3x + y - 66 = 0$ $\;\;\; \cdots \; (2)$

To determine the point $Q$ where the normal again meets the parabola, substitute $\;$ $y = 66 - 3x$ $\;$ from equation $(2)$ in equation $(1)$

$\therefore \;$ We have,

$\left(66 - 3x\right)^2 = 8x$

i.e. $\;$ $4356 + 9x^2 - 396x = 8x$

i.e. $\;$ $9x^2 - 404x + 4356 = 0$

Solving, we get, $\;$ $x = \dfrac{242}{9}$ $\;$ or $\;$ $x = 18$

When $\;$ $x = \dfrac{242}{9}$, $\;$ we have from equation $(2)$

$y = 66 - 3 \times \dfrac{242}{9} = \dfrac{-44}{3}$

$\therefore \;$ $Q = \left(\dfrac{242}{9}, \dfrac{-44}{3}\right)$

$\therefore \;$ $PQ = \sqrt{\left(18 - \dfrac{242}{9}\right)^2 + \left(12 + \dfrac{44}{3}\right)^2}$

i.e. $\;$ $PQ = \sqrt{\dfrac{6400}{81} + \dfrac{6400}{9}} = 80 \sqrt{\dfrac{1 + 9}{81}} = \dfrac{80 \sqrt{10}}{9}$

$\implies$ $9 PQ = 80 \sqrt{10}$

Hence proved.

Coordinate Geometry - Parabola

Find the equation of the normal to the parabola $y^2 = 4x$ which is perpendicular to the line $2x + 6y + 5 = 0$.

Equation of parabola: $\;$ $y^2 = 4x$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the standard equation of parabola $\;$ $y^2 = 4ax$ $\;$ gives

$4a = 4$ $\implies$ $a = 1$

Equation of line: $\;$ $2x + 6y + 5 = 0$ $\;\;\; \cdots \; (2)$

Slope of equation $(2)$ is $\;$ $m_1 = \dfrac{-2}{6} = \dfrac{-1}{3}$

The required normal is perpendicular to line $(2)$.

$\therefore \;$ Slope of normal $= m = \dfrac{-1}{m_1} = 3$

Let the equation of the required normal be $\;$ $y = mx - 2am - am^2$ $\;\;\; \cdots \; (3)$

Substituting the values of $a$ and $m$ in equation $(3)$ gives

$y = 3x - 2 \times 1 \times 3 - 1 \times 3^3$

i.e. $\;$ $3x - y - 33 = 0$ $\;\;\; \cdots \; (4)$

Equation $(4)$ is the equation of the required normal.

Coordinate Geometry - Parabola

Find the equation of the tangent to the parabola $y^2 = 16x$ which is parallel to the line $3x - 4y + 5 = 0$. Also find the point of contact.

Equation of parabola: $\;$ $y^2 = 16x$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the standard equation of parabola $\;$ $y^2 = 4ax$ $\;$ gives

$4a = 16$ $\implies$ $a = 4$

Equation of line: $\;$ $3x - 4y + 5 = 0$ $\;\;\; \cdots \; (2)$

Slope of equation $(2)$ is $\;$ $m = \dfrac{3}{4}$

The required tangent is parallel to line $(2)$.

$\therefore \;$ Slope of tangent $= m = \dfrac{3}{4}$

Let the equation of the required tangent be $\;$ $y = mx + \dfrac{a}{m}$ $\;\;\; \cdots \; (3)$

Substituting the values of $a$ and $m$ in equation $(3)$ gives

$y = \dfrac{3}{4} x + \dfrac{4}{3 / 4}$

i.e. $\;$ $y = \dfrac{3}{4} x + \dfrac{16}{3}$ $\;\;\; \cdots \; (4a)$

i.e. $\;$ $9x - 12y + 64 = 0$ $\;\;\; \cdots \; (4b)$

Equation $(4b)$ is the equation of the required tangent.

Substituting the value of $y$ from equation $(4a)$ in equation $(1)$ gives the point of contact.

$\therefore \;$ We have

$\left(\dfrac{3}{4} x + \dfrac{16}{3}\right)^2 = 16x$

i.e. $\;$ $\dfrac{9}{16} x^2 + \dfrac{256}{9} + 8x = 16x$

i.e. $\;$ $81 x^2 - 1152 x + 4096 = 0$

i.e. $\;$ $x = \dfrac{1152 \pm \sqrt{\left(-1152\right)^2 - 4 \times 81 \times 4096}}{2 \times 81}$

i.e. $\;$ $x = \dfrac{1152 \pm \sqrt{1327104 - 1327104}}{162}$

i.e. $\;$ $x = \dfrac{1152}{162} = \dfrac{64}{9}$

Substituting the value of $x$ in equation $(4a)$ gives

$y = \left(\dfrac{3}{4} \times \dfrac{64}{9}\right) + \dfrac{16}{3} = \dfrac{32}{3}$

$\therefore \;$ The point of contact is $\left(\dfrac{64}{9}, \dfrac{32}{3}\right)$.

Coordinate Geometry - Parabola

Find the value of $k$ so that the line $3x - y + k = 0$ may touch the parabola $y^2 = 24x$.

Given parabola: $\;$ $y^2 = 24x$ $\;\;\; \cdots \; (1)$

Given line: $\;$ $3x - y + k = 0$ $\;$ i.e. $\;$ $y = 3x + k$ $\;\;\; \cdots \; (2)$

Substituting the value of $y$ from equation $(2)$ in equation $(1)$ gives

$\left(3x + k\right)^2 = 24x$

i.e. $\;$ $9x^2 + 6kx + k^2 = 24x$

i.e. $\;$ $9x^2 + \left(6k - 24\right) x + k^2 = 0$ $\;\;\; \cdots \; (3)$

Since equation $(2)$ touches the parabola given by equation $(1)$, line $(2)$ meets the parabola in two coincident points.

i.e. $\;$ the discriminant of equation $(3)$ is equal to $0$ (zero).

i.e. $\;$ Discriminant $= \Delta = \left(6k - 24\right)^2 - 4 \times 9 \times k^2 = 0$

i.e. $\;$ $36 k^2 - 288 k + 576 - 36 k^2 = 0$

i.e. $\;$ $288 k = 576$

$\implies$ $k = 2$

Coordinate Geometry - Parabola

Show that the line $12y - 20x - 9 = 0$ touches the parabola $y^2 = 5x$.

Equation of parabola: $\;$ $y^2 = 5x$ $\;\;\; \cdots \; (1)$

Equation of line: $\;$ $12y - 20x - 9 = 0$ $\;$ i.e. $\;$ $y = \dfrac{20x + 9}{12}$ $\;\;\; \cdots \; (2)$

In view of equation $(2)$, equation $(1)$ becomes

$\left(\dfrac{5}{3} x + \dfrac{3}{4}\right)^2 = 5x$

i.e. $\;$ $\dfrac{25}{9} x^2 + \dfrac{9}{16} + \dfrac{5}{2} x = 5x$

i.e. $\;$ $\dfrac{25}{9} x^2 - \dfrac{5}{2} x + \dfrac{9}{16} = 0$ $\;\;\; \cdots \; (3)$

Discriminant of equation $(3)$ is

$\Delta = \left(\dfrac{-5}{2}\right)^2 - 4 \times \dfrac{25}{9} \times \dfrac{9}{16} = \dfrac{25}{4} - \dfrac{25}{4} = 0$

$\implies$ Roots of equation $(3)$ are equal.

$\implies$ Equation $(2)$ touches equation $(1)$

i.e. $\;$ equation $(2)$ is a tangent to the given parabola.

Coordinate Geometry - Parabola

Find the equations of the tangents to the parabola $y^2 + 12x = 0$, from the point $\left(3, 8\right)$.

Equation of parabola: $\;$ $y^2 + 12x = 0$ $\;$ i.e. $\;$ $y^2 = -12x$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the standard equation of parabola $\;$ $y^2 = 4ax$ $\;$ gives

$4a = -12$ $\implies$ $a = -3$

Let the equation of the required tangent be

$y = mx + \dfrac{a}{m}$ $\;\;\; \cdots \; (2a)$

Substituting the value of $\;$ $a$ $\;$ in equation $(2a)$ gives

$y = mx - \dfrac{3}{m}$ $\;\;\; \cdots \; (2b)$

Given: $\;$ Equation $(2b)$ is drawn from the point $\left(3, 8\right)$,

$\implies$ $8 = 3m - \dfrac{3}{m}$

i.e. $\;$ $3m^2 - 8m - 3 = 0$

i.e. $\;$ $\left(3m + 1\right) \left(m - 3\right) = 0$

i.e. $\;$ $m = \dfrac{-1}{3}$ $\;$ or $\;$ $m = 3$

Substituing the values of $m$ in equation $(2b)$ gives

when $\;$ $m = \dfrac{-1}{3}$

$y = \dfrac{-1}{3} x - \dfrac{3}{-1/3}$

i.e. $\;$ $3y = -x + 27$

i.e. $\;$ $x + 3y = 27$ $\;\;\; \cdots \; (3a)$

when $\;$ $m = 3$

$y = 3x - \dfrac{3}{3}$

i.e. $\;$ $3x - y = 1$ $\;\;\; \cdots \; (3b)$

Equations $(3a)$ and $(3b)$ are the required equations of tangents.

Coordinate Geometry - Parabola

A tangent to the parabola $\;$ $y^2 = 16x$ $\;$ makes an angle of $\;$ $60^\circ$ $\;$ with the axis. Find the point of contact.

Equation of parabola: $\;$ $y^2 = 16x$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the standard equation of parabola $\;$ $y^2 = 4ax$ $\;$ gives

$4a = 16$ $\implies$ $a = 4$

Let slope of tangent $= m$

Since the tangent makes an angle of $60^\circ$ with the axis,

$\implies$ $m = \tan 60^\circ = \sqrt{3}$

Point of contact of the tangent with the parabola $= \left(\dfrac{a}{m^2}, \dfrac{2a}{m}\right) = \left(\dfrac{4}{\left(\sqrt{3}\right)^2}, \dfrac{2 \times 4}{\sqrt{3}}\right)$

$\therefore \;$ The point of contact $= \left(\dfrac{4}{3}, \dfrac{8}{\sqrt{3}}\right)$

Coordinate Geometry - Parabola

Find the tangent to the parabola $\;$ $y^2 = 16x$ $\;$ making an angle of $\;$ $45^\circ$ $\;$ with the $X$ axis.

Let the equation of the tangent be

$y = mx + \dfrac{a}{m}$ $\;\;\; \cdots \; (1)$ $\;\;\;$ where $m$ is the slope of the tangent

The tangent makes an angle of $45^\circ$ with the $X$ axis.

$\therefore \;$ $m = \tan 45^\circ = 1$

Given: $\;$ Equation of parabola: $\;\;\;$ $y^2 = 16x$ $\;\;\; \cdots \; (2)$

Comparing equation $(2)$ with the standard equation of parabola $\;$ $y^2 = 4ax$ $\;$ gives

$4a = 16$ $\implies$ $a = 4$

Substituting the values of $m$ and $a$ in equation $(1)$ gives

$y = 1 \times x + \dfrac{4}{1}$

i.e. $\;$ $y = x + 4$ $\;\;\; \cdots \; (3)$

Equation $(3)$ is the required equation of tangent.

Coordinate Geometry - Parabola

Find the equation of the parabola whose axis is parallel to the $X$ axis and the parabola passes through $\left(3, 3\right)$, $\left(6, 5\right)$ and $\left(6, -3\right)$.

The axis of the required parabola is parallel to the $X$ axis.

$\therefore \;$ Let the equation of the parabola be

$x = Ay^2 + By + C$ $\;\;\; \cdots \; (1)$

Since equation $(1)$ passes through the points $\left(3, 3\right)$, $\left(6, 5\right)$ and $\left(6, -3\right)$, we have

$3 = 9A + 3B + C$ $\;\;\; \cdots \; (2a)$

$6 = 25 A + 5B + C$ $\;\;\; \cdots \; (2b)$

$6 = 9A - 3B + C$ $\;\;\; \cdots \; (2c)$

Solving equations $(2a)$ and $(2c)$ simultaneously gives

$- 3 = 6 B$ $\implies$ $B = \dfrac{-1}{2}$

Solving equations $(2a)$ and $(2b)$ simultaneously gives

$3 = 16A + 2B$ $\;\;\; \cdots \; (3a)$

Substituting the value of $B$ in equation $(3a)$ gives

$3 = 16A + 2 \times \left(\dfrac{-1}{2}\right)$

i.e. $\;$ $4 = 16 A$ $\implies$ $A = \dfrac{1}{4}$

Substituting the values of $A$ and $B$ in equation $(2a)$ gives

$3 = 9 \times \dfrac{1}{4} + 3 \times \left(\dfrac{-1}{2}\right) + C$

i.e. $\;$ $C = 3 - \dfrac{9}{4} + \dfrac{3}{2} = \dfrac{9}{4}$

Substituting the values of $A$, $B$ and $C$ in equation $(1)$ gives

$x = \dfrac{1}{4} y^2 - \dfrac{1}{2} y + \dfrac{9}{4}$

i.e. $\;$ $y^2 - 2y - 4x + 9 = 0$ $\;\;\; \cdots \; (4)$

Equation $(4)$ is the required equation of the parabola.

Coordinate Geometry - Parabola

Find the equation of the parabola whose vertex is at $\left(3, -2\right)$ and the focus at $\left(6, 2\right)$.

Given: $\;$ Vertex $= V = \left(3, -2\right)$, $\;\;\;$ Focus $= F = \left(6, 2\right)$

The axis is the line joining the vertex and the focus.

$\therefore \;$ Slope of axis $= \dfrac{2 + 2}{6 - 3} = \dfrac{4}{3}$

The directrix is the line perpendicular to the axis and passing through the vertex.

$\therefore \;$ Slope of directrix $= \dfrac{-3}{4}$

Let $\;$ $Z \left(x_1, y_1\right)$ $\;$ be the point of intersection of the axis and the directrix.

Then, vertex $\;$ $V \left(3, -2\right)$ $\;$ is the midpoint of the segment $\;$ $ZF$ $\;$ where $\;$ $Z = \left(x_1, y_1\right)$ $\;$ and $\;$ $F = \left(6, 2\right)$

$\therefore \;$ By midpoint formula,

$3 = \dfrac{x_1 + 6}{2}$ $\implies$ $x_1 = 0$

and $\;$ $-2 = \dfrac{y_1 + 2}{2}$ $\implies$ $y_1 = -6$

$\therefore \;$ $Z \left(x_1, y_1\right) = \left(0, -6\right)$

$\therefore \;$ Equation of directrix $\;$ (with slope $\dfrac{-3}{4}$ and passing through $Z$) $\;$ is

$y + 6 = \dfrac{-3}{4} \left(x - 0\right)$

i.e. $\;$ $3x + 4y + 24 = 0$

Let $\;$ $P \left(x, y\right)$ $\;$ be any point on the required parabola.

Then, by definition,

distance of $P$ from focus $=$ distance of $P$ from the directrix

i.e. $\;$ $\sqrt{\left(x - 6\right)^2 + \left(y - 2\right)^2} = \left|\dfrac{3x + 4y + 24}{\sqrt{3^2 + 4^2}}\right|$

i.e. $\;$ $x^2 + y^2 - 12x - 4y + 36 + 4 = \dfrac{9x^2 + 16y^2 + 576 + 24xy + 144x + 192y}{5}$

i.e. $\;$ $5x^2 + 5y^2 - 60x - 20y + 200 = 9x^2 + 16y^2 + 24xy + 144x + 192y + 576$

i.e. $\;$ $4x^2 + 11y^2 - 24xy + 204x + 212y + 376 = 0$ $\;\;\; \cdots \; (1)$

Equation $(1)$ is the equation of the required parabola.

Coordinate Geometry - Parabola

Find the equation of the parabola whose focus is at $\left(-2, -1\right)$ and the latus rectum joins the points $\left(-2, 2\right)$ and $\left(-2, -4\right)$.

Given: $\;$ Latus rectum joins the points $\;$ $\left(-2, 2\right)$ $\;$ and $\;$ $\left(-2, -4\right)$

Length of latus rectus $= \sqrt{\left(-2 + 2\right)^2 + \left(2 + 4\right)^2} = 6$

But length of latus rectum $= 4a$

i.e. $\;$ $4a = 6$ $\implies$ $a = \dfrac{6}{4} = \dfrac{3}{2}$

Given: $\;$ Focus $= \left(-2, -1\right)$

Let the vertex of the required parabola be $= \left(\alpha, -1\right)$

Then,

Distance between focus and vertex $= a$

i.e. $\;$ $\sqrt{\left(\alpha + 2\right)^2 + \left(-1 + 1\right)^2} = \dfrac{3}{2}$

i.e. $\;$ $\alpha + 2 = \dfrac{3}{2}$ $\implies$ $\alpha = \dfrac{-1}{2}$

$\therefore \;$ The vertex of the required parabola is $= \left(h, k\right) = \left(\dfrac{-1}{2}, -1\right)$

Let the equation of the required parabola be

$\left(y - k\right)^2 = \pm 4 a \left(x - h\right)$

i.e. $\;$ $\left(y + 1\right)^2 = \pm 6 \left(x + \dfrac{1}{2}\right)$

i.e. $\;$ $y^2 + 2y + 1 = 6x + 3$ $\;\;$ or $\;\;$ $y^2 + 2y + 1 = -6x - 3$

i.e. $\;$ $y^2 + 2y - 6x -2 = 0$ $\;\;\; \cdots \; (1a)$

or $\;$ $y^2 + 2y + 6x + 4 = 0$ $\;\;\; \cdots \; (1b)$

Equations $(1a)$ and $(1b)$ are the required equations of the parabola.

Coordinate Geometry - Parabola

Find the equation of the parabola whose focus is at $\left(1,1\right)$ and the directrix is $x - y = 3$.

Given: $\;$ Equation of directrix of parabola is $\;\;$ $x - y = 3$ $\;\;$ i.e. $\;$ $x - y -3 = 0$

Given: $\;$ Focus of parabola is at $\;$ $\left(1, 1\right)$

Let $P \left(x, y\right)$ be any point on the parabola.

Then, by definition of parabola,

Distance of $P$ from the focus $=$ distance of point $P$ from the directrix

i.e. $\;$ $\sqrt{\left(x - 1\right)^2 + \left(y - 1\right)^2} = \left|\dfrac{x - y - 3}{\sqrt{1^2 + \left(-1\right)^2}}\right|$

i.e. $\;$ $x^2 - 2x + 1 + y^2 - 2y + 1 = \dfrac{x^2 + y^2 + 9 - 2xy - 6x + 6y}{2}$

i.e. $\;$ $2x^2 - 4x + 2y^2 - 4y + 4 = x^2 + y^2 - 6x + 6y - 2xy + 9$

i.e. $\;$ $x^2 + y^2 + 2xy + 2x - 10 y - 5 = 0$ $\;\;\; \cdots \; (1)$

Equation $(1)$ is the required equation of the parabola.

Coordinate Geometry - Parabola

Find the vertex, axis, focus and directrix of the parabola $\;$ $5y^2 - 20y - 16x + 96 = 0$.

Equation of given parabola: $\;$ $5y^2 - 20y - 16x + 96 = 0$

i.e. $\;$ $y^2 - 4y - \dfrac{16}{5}x + \dfrac{96}{5} = 0$

i.e. $\;$ $\left(y^2 - 4y + 4\right) = \dfrac{16}{5} x - \dfrac{96}{5} + 4$

i.e. $\;$ $\left(y - 2\right)^2 = \dfrac{16}{5} x - \dfrac{76}{5}$

i.e. $\;$ $\left(y - 2\right)^2 = \dfrac{16}{5} \left(x - \dfrac{76}{16}\right)$

i.e. $\;$ $\left(y - 2\right)^2 = \dfrac{16}{5} \left(x - \dfrac{19}{4}\right)$ $\;\;\; \cdots \; (1)$

To shift the origin to the point $\left(\dfrac{19}{4}, 2\right)$,

let $\;$ $x - \dfrac{19}{4} = X$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $y - 2 = Y$ $\;\;\; \cdots \; (2b)$

In view of equations $(2a)$ and $(2b)$, equation $(1)$ becomes

$Y^2 = \dfrac{16}{5} X$ $\;\;\; \cdots \; (3)$

1. Vertex of equation $(3)$ is $\;$ $\left(0, 0\right)$
   
   i.e. $\;$ $X = 0, \;\;\; Y = 0$
   
   When $\;$ $X = 0$
   
   $\implies$ $x - \dfrac{19}{4} = 0$ $\;$ i.e. $\;$ $x = \dfrac{19}{4}$ $\;\;\;$ [by equation $(2a)$]
   
   When $\;$ $Y = 0$
   
   $\implies$ $y - 2 = 0$ $\;$ i.e. $\;$ $y = 2$ $\;\;\;$ [by equation $(2b)$]
   
   $\therefore \;$ Vertex of equation $(1)$ is $\;$ $\left(\dfrac{19}{4}, 2\right)$
   
   
2. Axis of equation $(3)$ is $\;\;$ $Y = 0$
   
   $\therefore \;$ Axis of equation $(1)$ is $\;\;\;$ $y - 2 = 0$ $\;\;\;$ [by equation $(2b)$]
   
   i.e. $\;$ $y = 2$
   
   
3. Comparing equation $(3)$ with the standard equation of parabola $\;\;$ $Y^2 = 4a X$ $\;\;$ gives
   
   $4a = \dfrac{16}{5}$ $\implies$ $a = \dfrac{4}{5}$
   
   $\therefore \;$ Focus of equation $(3)$ is $\;\;\;$ $\left(a, 0\right) = \left(\dfrac{4}{5}, 0\right)$
   
   $\implies$ $X = \dfrac{4}{5}, \;\;\; Y = 0$
   
   When $\;$ $X = \dfrac{4}{5}$
   
   $\implies$ $x - \dfrac{19}{4} = \dfrac{4}{5}$ $\;$ i.e. $\;$ $x = \dfrac{111}{20}$ $\;\;\;$ [by equation $(2a)$]
   
   When $\;$ $Y = 0$
   
   $\implies$ $y - 2 = 0$ $\;$ i.e. $\;$ $y = 2$ $\;\;\;$ [by equation $(2b)$]
   
   $\therefore \;$ Focus of equation $(1)$ is $\;\;\;$ $\left(\dfrac{111}{20}, 2\right)$
   
   
4. Directrix of equation $(3)$ is $\;\;\;$ $X = -a$
   
   i.e. $\;$ $X = \dfrac{-4}{5}$
   
   i.e. $x - \dfrac{19}{4} = \dfrac{-4}{5}$ $\;\;$ i.e. $\;\;$ $x = \dfrac{79}{20}$ $\;\;\;$ [by equation $(2a)$]
   
   $\therefore \;$ Directrix of equation $(1)$ is $\;\;\;$ $20 x - 79 = 0$

Coordinate Geometry - Parabola

Find the vertex, axis, focus and directrix of the parabola $\;$ $x^2 + 2gx + 2fy + c = 0$.

Equation of given parabola: $\;$ $x^2 + 2gx + 2fy +c = 0$

i.e. $\;$ $\left(x^2 + 2gx+ g^2\right) + 2fy + c - g^2 = 0$

i.e. $\;$ $\left(x + g\right)^2 = -2fy + g^2 - c$

i.e. $\;$ $\left(x + g\right)^2 = -2f \left(y - \dfrac{g^2 - c}{2f}\right)$ $\;\;\; \cdots \; (1)$

To shift the origin to the point $\left(-g, \dfrac{g^2 - c}{2f}\right)$,

let $\;$ $x + g = X$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $y - \dfrac{g^2 - c}{2f} = Y$ $\;\;\; \cdots \; (2b)$

In view of equations $(2a)$ and $(2b)$, equation $(1)$ becomes

$X^2 = -2fY$ $\;\;\; \cdots \; (3)$

1. Vertex of equation $(3)$ is $\;$ $\left(0, 0\right)$
   
   i.e. $\;$ $X = 0, \;\;\; Y = 0$
   
   When $\;$ $X = 0$
   
   $\implies$ $x + g = 0$ $\;$ i.e. $\;$ $x = -g$ $\;\;\;$ [by equation $(2a)$]
   
   When $\;$ $Y = 0$
   
   $\implies$ $y - \dfrac{g^2 - c}{2f} = 0$ $\;$ i.e. $\;$ $y = \dfrac{g^2 - c}{2f}$ $\;\;\;$ [by equation $(2b)$]
   
   $\therefore \;$ Vertex of equation $(1)$ is $\;$ $\left(-g, \dfrac{g^2 - c}{2f}\right)$
   
   
2. Axis of equation $(3)$ is $\;\;$ $X = 0$
   
   $\therefore \;$ Axis of equation $(1)$ is $\;\;\;$ $x + g = 0$ $\;\;\;$ [by equation $(2a)$]
   
   i.e. $\;$ $x = -g$
   
   
3. Comparing equation $(3)$ with the standard equation of parabola $\;\;$ $X^2 = -4a Y$ $\;\;$ gives
   
   $4a = 2f$ $\implies$ $a = \dfrac{f}{2}$
   
   $\therefore \;$ Focus of equation $(3)$ is $\;\;\;$ $\left(0, -a\right) = \left(0, \dfrac{-f}{2}\right)$
   
   $\implies$ $X = 0, \;\;\; Y = \dfrac{-f}{2}$
   
   When $\;$ $X = 0$
   
   $\implies$ $x + g = 0$ $\;$ i.e. $\;$ $x = -g$ $\;\;\;$ [by equation $(2a)$]
   
   When $\;$ $Y = \dfrac{-f}{2}$
   
   $\implies$ $y - \dfrac{g^2 - c}{2f} = \dfrac{-f}{2}$ $\;$ i.e. $\;$ $y = \dfrac{g^2 - f^2 - c}{2f}$ $\;\;\;$ [by equation $(2b)$]
   
   $\therefore \;$ Focus of equation $(1)$ is $\;\;\;$ $\left(-g, \dfrac{g^2 - f^2 - c}{2f}\right)$
   
   
4. Directrix of equation $(3)$ is $\;\;\;$ $Y = a$
   
   i.e. $\;$ $Y = \dfrac{f}{2}$
   
   i.e. $y - \dfrac{g^2 - c}{2f} = \dfrac{f}{2}$ $\;\;\;$ [by equation $(2b)$]
   
   i.e. $\;$ $y= \dfrac{g^2 - f^2 - c}{2f}$
   
   $\therefore \;$ Directrix of equation $(1)$ is $\;\;\;$ $y = \dfrac{g^2 - f^2 - c}{2f}$