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 $\left(\dfrac{64}{9}, \dfrac{32}{3}\right)$.