Question

If $F_1$ and $F_2$ are the feet of the perpendiculars from the foci $S_1$ and $S_2$ of an ellipse $\frac{x^2}{5}+\frac{y^2}{3}=1$ on the tangent at any point $P$ on the ellipse, then $\left(S_1{F}_1\right)\left(S_2{F}_2\right)$ is equal to

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is B $3$

Given, $\frac{x^2}{5}+\frac{y^2}{3}=1$

We know $S_1{F}_1\times S_2{F}_2=b^2$

$\therefore S_1{F}_1\times S_2{F}_2=3$