Question

If $P(\alpha ,\beta )$ is a point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with foci $S$ and $S^{\prime }$ and eccentricity $e,$ then area of $\Delta SP{S}^{\prime }$ is

• A. $ae\sqrt{a^2-{\alpha }^2}$
• B. $be\sqrt{b^2-{\alpha }^2}$
• C. $ae\sqrt{b^2-{\alpha }^2}$
• D. $be\sqrt{a^2-{\alpha }^2}$

Answer 1

The correct option is D $be\sqrt{a^2-{\alpha }^2}$

Given, $P(\alpha ,\beta )$ is a point on the ellipse equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
On substituting point in the ellipse equation.
$\Rightarrow \frac{{\alpha }^2}{a^2}+\frac{{\beta }^2}{b^2}=1$
$\Rightarrow \beta =\frac{b}{a}\times \sqrt{a^2-{\alpha }^2}...\left(A\right)$

Area of triangle

$S^{\prime }PS=\frac{1}{2}\times Base\times Height$

$=\frac{1}{2}\times \left(S{S}^{\prime }\right)\times \beta$

(Since, $S=(ae,0),S^{\prime }=(-ae,0)\Rightarrow S{S}^{\prime }=2ae$)

$=\frac{1}{2}\times \left(2ae\right)\times \frac{b}{a}\times \sqrt{a^2-{\alpha }^2}$

$=be\times \sqrt{a^2-{\alpha }^2}$