Question

If the equation of ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ (where $a>b$ ), then the length of latusrectum is

• A. $\frac{2 a^2}{b}$
• B. $\frac{2 b^2}{a}$
• C. $\frac{b^2}{a}$
• D. $\frac{a^2}{b}$

Answer 1

Answer: (B)

To find the length of the latusrectum of the ellipse
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \text {. }$
Let the length of $A F_2$ be $I$.

Then, the coordinates of $A$ are $(c, I)$, i.e., (ae, I)
Since, $A$ lies on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,
we have $ \frac{(a e)^2}{a^2}+\frac{l^2}{b^2} =1 $
$ \Rightarrow l^2 =b^2\left(1-e^2\right) $
$ \text { But } e^2 =\frac{c^2}{a^2}=\frac{a^2-b^2}{a^2}=1-\frac{b^2}{a^2} $
$\Rightarrow \frac{b^2}{a^2} =1-e^2$
Therefore, $I^2=\frac{b^4}{a^2}$, i.e., $I=\frac{b^2}{a}$
Since, the ellipse is symmetric with respect to $Y$-axis ( of course, it is symmetric w.r.t both the coordinate axes),
$A F_2=F_2 B$ and so length of the latusrectum is $\frac{2 b^2}{a}$.