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

The correct option is A $2b(be+a{\sin }^{-1}e)$

Required area is
$A=4\underset{0}{\overset{ae}{\int }}ydx$
$A=4\underset{0}{\overset{ae}{\int }}\sqrt{b^2-\frac{b^2{x}^2}{a^2}}dx$

Let $x=a\sin y$
$\Rightarrow dx=a\cos ydy$
$A=4\underset{0}{\overset{{\sin }^{-1}e}{\int }}\sqrt{b^2-b^2{\sin }^{2}y}\left(a\cos ydy\right)$
$\Rightarrow A=4\underset{0}{\overset{{\sin }^{-1}e}{\int }}ab{\cos }^{2}ydy$
$\Rightarrow A=4ab\underset{0}{\overset{{\sin }^{-1}e}{\int }}\frac{1+\cos 2y}{2}dy$
$\Rightarrow A=2ab[{\sin }^{-1}e+\frac{\sin \left(2{\sin }^{-1}e\right)}{2}]$

$\Rightarrow A=2ab[{\sin }^{-1}e+\sin ({\sin }^{-1}e\left)\cos \right({\sin }^{-1}e\left)\right]$
$\Rightarrow A=2ab[{\sin }^{-1}e+e\sqrt{1-e^2}]$
$\Rightarrow A=2ab[{\sin }^{-1}e+e\sqrt{1-1+\frac{b^2}{a^2}}]$
$\Rightarrow A=2ab[{\sin }^{-1}e+\frac{eb}{a}]$
$\Rightarrow A=2b[a{\sin }^{-1}e+eb]$