Question

If $F_1$ and $F_2$ are the feet of perpendiculars from 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 $\left(S_1{F}_1\right)$ $\left(S_2{F}_2\right)=$

• A. $8$
• B. $5$
• C. $3$
• D. $9$

Answer 1

Answer: (C)

$3$

Given,

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

$a^2=5,b^2=3$

$e$=eccentricity=$\sqrt{1-(\frac{b}{a}{)}^2}$

$=>e=\sqrt{\frac{2}{5}}$

Now,

tangent at any point $p$,we can take point $p$ at end of major axis.

$S_{1}F=a-ae$

$S_{2}F=a+ae$

Now,

$\left(S_{1}F\right)\left(S_{2}F\right)=(a+ae)(a-ae)$

$=>a^2-a^2{e}^2$

$=>5-5\times \frac{2}{5}$

$=>3$