Find the equations of directrices, latus rectum and length of latus rectum for the ellipse $3x^2 + 2y^2 - 30 x - 4y + 23 = 0$
Equation of given ellipse is: $\;$ $3x^2 + 2y^2 - 30 x - 4y + 23 = 0$
i.e. $\;$ $3 \left(x^2 - 10x\right) + 2 \left(y^2 - 2y\right) = -23$
i.e. $\;$ $3 \left[\left(x^2 - 10x + 25\right) - 25\right] + 2 \left[\left(y^2 - 2y + 1\right) -1\right] = -23$
i.e. $\;$ $3 \left(x - 5\right)^2 + 2 \left(y - 1\right)^2 - 75 - 2 = - 23$
i.e. $\;$ $3 \left(x - 5\right)^2 + 2 \left(y - 1\right)^2 = 54$
i.e. $\;$ $\dfrac{\left(x - 5\right)^2}{54 / 3} + \dfrac{\left(y - 1\right)^2}{54 / 2} = 1$
i.e. $\;$ $\dfrac{\left(x - 5\right)^2}{18} + \dfrac{\left(y - 1\right)^2}{27} = 1$ $\;\;\; \cdots \; (1)$
Let $\;$ $X = x - 5$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $Y = y - 1$ $\;\;\; \cdots \; (2b)$
Then in view of equations $(2a)$ and $(2b)$, equation $(1)$ becomes
$\dfrac{X^2}{18} + \dfrac{Y^2}{27} = 1$ $\;\;\; \cdots \; (3)$
Comparing equation $(3)$ with the standard equation of ellipse $\;$ $\dfrac{X^2}{b^2} + \dfrac{Y^2}{a^2} = 1$ $\;$ (for $b < a$) $\;$ gives
$b^2 = 18$ $\;\;\; \cdots \; (4a)$; $\;$ $a^2 = 27 \implies a = 3 \sqrt{3}$ $\;\;\; \cdots \; (4b)$
Now, $e^2 = 1 - \dfrac{b^2}{a^2} = 1 - \dfrac{18}{27} = \dfrac{9}{27}$ $\implies$ $e = \dfrac{1}{\sqrt{3}}$ $\;\;\; \cdots \; (4c)$
Equations of directrices are $\;$ $Y = \pm \dfrac{a}{e}$ $\;\;\; \cdots \; (5)$
$\therefore$ $\;$ In view of equations $(2b)$, $(4b)$ and $(4c)$, equation $(5)$ becomes
$y - 1 = \pm \dfrac{3 \sqrt{3}}{3}$
i.e. $\;$ $y - 1 = \pm 9$
i.e. $\;$ $y = 10$ $\;\;\; \cdots \; (5a)$ $\;$ and $\;$ $y = -8$ $\;\;\; \cdots \; (5b)$
Equations of latus rectum are: $\;$ $Y = \pm ae$ $\;\;\; \cdots \; (6)$
$\therefore$ $\;$ In view of equations $(2b)$, $(4b)$ and $(4c)$, equation $(6)$ becomes
$y - 1 = \pm 3 \sqrt{3 \times \dfrac{1}{\sqrt{3}}}$
i.e. $\;$ $y - 1 = \pm 3$
i.e. $\;$ $y = 4$ $\;\;\; \cdots \; (6a)$ $\;$ and $y = -2$ $\;\;\; \cdots \; (6b)$
Length of latus rectum $= \dfrac{2b^2}{a} = \dfrac{2 \times 18}{3 \sqrt{3}} = 4 \sqrt{3}$ $\;$ [from equations $(4a)$ and $(4b)$]