Question

Find the area of greater rectangle that can be inscribed in an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

• A. $-8ab\sin 2\theta$
• B. $-6ab\sin 2\theta$
• C. $8ab\sin 2\theta$
• D. $2ab\sin 2\theta$

Answer 1

The correct option is A $-8ab\sin 2\theta$

Given that

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

$\Rightarrow \frac{y^2}{b^2}=1-\frac{x^2}{a^2}$

$\Rightarrow y^2=b^2\left(\frac{a^2-x^2}{a^2}\right)$

$\Rightarrow y^2=\frac{b^2}{a^2}\left(a^2-x^2\right)$

$\Rightarrow y=\sqrt{\frac{b^2}{a^2}\left(a^2-x^2\right)}$

$\Rightarrow y=\frac{b}{a}\sqrt{a^2-x^2}$

Required axis$={\int }_0^{a}4\frac{b}{a}\sqrt{a^2-x^2}dx$

$=4\frac{b}{a}{\int }_0^a\sqrt{a^2-x^2}dx$

$=-8ab\sin 2\theta$

Hence, the option $\left(A\right)$ is correct answer.