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. $\frac{2a^2}{b}$
• B. $\frac{2b^2}{a}$
• 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 $\frac{(ae{)}^2}{a^2}+\frac{l^2}{b^2}=1$
$\Rightarrow l^2=b^2\left(1-e^2\right)$
$\text{ But }{e}^2=\frac{c^2}{a^2}=\frac{a^2-b^2}{a^2}=1-\frac{b^2}{a^2}$
$\Rightarrow \frac{b^2}{a^2}=1-e^2$
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 $\frac{2b^2}{a}$.