Question

Consider an ellipse, whose centre is at the origin and its major axis is along the x-axis. If its eccentricity is $\frac{3}{5}$ and the distance between its foci is $6$, then the area (in sq. units) of the quadrilateral inscribed in the ellipse, with the vertices as the vertices of the ellipse, is.

• A. $80$
• B. $8$
• C. $40$
• D. $32$

Answer 1

The correct option is C $40$

The required area is in the shape of kite.

Area of kite $=\frac{1}{2}\times$ ( Product of its diagonal )

$\Rightarrow$ Eccentricity $\left(e\right)=\frac{3}{5}$

$\Rightarrow$ Distance between foci $=6$

$\therefore$ $2ae=6$

$\Rightarrow$ $ae=3$

$\Rightarrow$ $a\times \frac{3}{5}=3$

$\therefore$ $a=5$

Now,

$\Rightarrow$ $b^2=a^2-(ae{)}^2$

$\Rightarrow$ $b^2=(5{)}^2-(3{)}^2$

$\Rightarrow$ $b^2=25-9$

$\Rightarrow$ $b^2=16$

$\therefore$ $b=4$

$\Rightarrow$ Length of major axis $\left(A{A}^{\prime }\right)=2a$

$=2\times 5$

$=10$

$\Rightarrow$ Length of minor axis $\left(B{B}^{\prime }\right)=2b$

$=2\times 4$

$=8$

$\Rightarrow$ Required are $=$ Area of kite $AB{A}^{\prime }{B}^{\prime }$

$=\frac{1}{2}\times \left(A{A}^{\prime }\right)\times \left(B{B}^{\prime }\right)$

$=\frac{1}{2}\times 10\times 8$

$=40$