Find the value of $a$ for which the ellipse $\frac{x^{2}}{y^{2}} + \frac{y^{2}}{b^{2}} = 1, (a > b)$ , if the extremities of the latus rectum of the ellipse having positive ordinates lie on the parabola $x^{2} = - 2 (y - 2)$ .

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Solution

Given $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
Extermities $(\pm a e, \frac{b^{2}}{a})$
$x^{2} = 2 (y - 2)$
$\Rightarrow a^{2} e^{2} = 2 (\frac{b^{2}}{a} - 2)$
we know $b^{2} = a^{2} (1 - e^{2})$
$\Rightarrow a^{2} e^{2} = 2 (\frac{a^{2} (1 - e^{2})}{a} - 2)$
$a^{2} e^{2} = 2 (a - a e^{2} - 2)$
$\Rightarrow a^{2} e^{2} = 2 a - 2 a e^{2} - y$
$\Rightarrow a^{2} e^{2} - 2 a e^{2} + 2 a - 4 = 0$
$\Rightarrow a e^{2} (a - 2) + 2 (a - 2) = 0$
$\Rightarrow (a e^{2} + 2) (a - 2) = 0$
$∴ a = 2$

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