Question

Extremities of the latus rectum of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$ having a given major axis 2a lies on

• A. $x^2=a(a-y)$
• B. $y^2=a(a+x)$
• C. $y^2=a(a-x)$
• D. None of these

Answer 1

Answer: (A)

Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
be an ellipse with eccentricity $e$,
then ends of L.R are $\left(ae,\pm \frac{b^2}{a}\right)$
and $\left(-ae,\pm \frac{b^2}{a}\right)$
Let $x_1=ae$ and $y_1=b^2/a$
$\Rightarrow y_1=\frac{a^2}{a}\left(1-e^2\right)$
$\Rightarrow \frac{y_1}{a}=1-{\left(\frac{x_1}{a}\right)}^2$
$\Rightarrow x^2=a(a-y)$