Question

The distance between the polars of the foci of the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$

• A. $\frac{25}{9}$
• B. $\frac{25}{16}$
• C. $\frac{25}{2}$
• D. $\frac{25}{3}$

Answer 1

Answer: (C)

$\frac{25}{2}$

Ellipse:$\frac{x^2}{25}+\frac{y^2}{9}=1---\left(i\right)$

eccentricity: $e=\sqrt{1-\frac{9}{25}}=\frac{4}{5}$

Foci: $F(4,0)$ and $F^{\prime }(-4,0)$

Polar of $F\Rightarrow T$ at $F$

$\Rightarrow \frac{4x}{25}+\frac{0}{9}=1$

$\Rightarrow x=\frac{25}{4}---\left(ii\right)$

Polar of $F^{\prime }\Rightarrow T$ at $F^{\prime }$

$\Rightarrow \frac{-4x}{25}+0=1$

$\Rightarrow x=\frac{-25}{4}---\left(iii\right)$

Perpendicular distance between $\left(ii\right)$ and $\left(iii\right)$,

$\Rightarrow \frac{25}{4}-(-\frac{25}{4})$

$\Rightarrow \frac{50}{4}$

$\Rightarrow \frac{25}{2}$