Question

Area bounded between two latus-rectum of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1;a>b$ is ________.
(where, $e$ is eccentricity of the ellipse).

• A. $2b(be+a{\sin }^{-1}e)$
• B. $8b(be+a{\sin }^{-1}e)$
• C. $b(be+a{\sin }^{-1}e)$
• D. $4b(be+a{\sin }^{-1}e)$

Answer 1

Answer: (A)

$2b(be+a{\sin }^{-1}e)$

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1⟹y=\sqrt{b^2-\frac{b^2{x}^2}{a^2}}$

As the given ellipse is symmetric, the area bound between the latus rectum can be divided into 4 equal parts using the major axis, the minor axis and the 2 latus recta.

The area of 1 such part = ${\int }_0^{ae}ydx$

$={\int }_0^{ae}\sqrt{b^2-\frac{b^2{x}^2}{a^2}}dx$

Substitute $x=a\sin y⟹dx=a\cos ydy$

$={\int }_0^{{\sin }^{-1}e}\sqrt{b^2-b^2{\sin }^{2}y}\left(a\cos ydy\right)$

$={\int }_0^{{\sin }^{-1}e}b\cos y\left(a\cos ydy\right)$

$=ab{\int }_0^{{\sin }^{-1}e}\frac{1+\cos 2y}{2}dy$ ($\because \cos 2x=2{\cos }^{2}x-1$)

$=\frac{ab}{2}({\sin }^{-1}e+\frac{\sin \left(2{\sin }^{-1}e\right)}{2})$

$=\frac{ab}{2}({\sin }^{-1}e+\sin ({\sin }^{-1}e\left)\cos \right({\sin }^{-1}e\left)\right)$

$=\frac{ab}{2}({\sin }^{-1}e+e\sqrt{1-e^2})$

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

$=\frac{ab}{2}({\sin }^{-1}e+e\frac{b}{a})$

Area of 4 such parts $=4\frac{ab}{2}({\sin }^{-1}e+e\frac{b}{a})$

$=2b(a{\sin }^{-1}e+be)$

This is the required solution.