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Q.
Extremities of the latus rectum of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \quad(a>b)$ having a given major axis 2a lies on
Conic Sections
Solution:
Let $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
be an ellipse with eccentricity $e$,
then ends of L.R are $\left(a e, \pm \frac{b^{2}}{a}\right)$
and $\left(-a e, \pm \frac{b^{2}}{a}\right)$
Let $x_{1}=a e$ and $y_{1}=b^{2} / a$
$\Rightarrow y _{1}=\frac{ a ^{2}}{ a }\left(1- e ^{2}\right) $
$\Rightarrow \frac{ y _{1}}{ a }=1-\left(\frac{ x _{1}}{ a }\right)^{2}$
$\Rightarrow x^{2}=a(a-y)$