Question

Area of the rectangle formed by the ends of latusrectum of the ellipse $4x^2+9y^2=144$ is

• A. $\frac{32\sqrt{5}}{3}$
• B. $\frac{64\sqrt{5}}{3}$
• C. $\frac{16\sqrt{5}}{3}$
• D. $\frac{32\sqrt{3}}{5}$

Answer 1

The correct option is B $\frac{64\sqrt{5}}{3}$

Given ellipse $4x^2+9y^2=144$

This can be re-written as $\frac{x^2}{36}+\frac{y^2}{16}=1$

Comparing it to the general form of ellipse we get $a=6,b=4$

Now eccentricity e is given by $e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{16}{36}}=\frac{2\sqrt{5}}{6}$

Now the rectangle is formed by the ends of the latus rectum.

The ends of latus rectum are given by $\{(ae,\frac{b^2}{a}),(-ae,\frac{b^2}{a}),(-ae,-\frac{b^2}{a}),(ae,-\frac{b^2}{a})\}$

Putting values of $a,b,e$ we gethe ends of latus rectum to be

$\{(2\sqrt{5},\frac{8}{3}),(-2\sqrt{5},\frac{8}{3}),(-2\sqrt{5},-\frac{8}{3}),(2\sqrt{5},-\frac{8}{3})\}$

Now to find the area of rectangle, find distances between the two points where area $A=l\times b^{\prime }$

$\Rightarrow l=\sqrt{{\left(4\sqrt{5}\right)}^2}=\left(4\sqrt{5}\right),b^{\prime }=\sqrt{{\left(\frac{16}{3}\right)}^2}=\frac{16}{3}$

Thus Area $\Rightarrow A=\left(4\sqrt{5}\right)\times \frac{16}{3}=64\frac{\sqrt{5}}{3}$