Question

The maximum area of the triangle inscribed in the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is

• A. $ab$
• B. $\sqrt{3ab}$
• C. $\frac{3\sqrt{3}}{4}ab$
• D. $\frac{ab}{4}$

Answer 1

The correct option is C $\frac{3\sqrt{3}}{4}ab$

let $ABC$ an isosceles triangle. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\left(given\right)$

C lies on the end of the major

axis,$AC=BC$

let $A(a\cos \theta ,b\sin \theta )$

$B(a\sin \theta ,-b\sin \theta )$

Let A is the triangle of the inscribed triangle

,then $A=\frac{(AB\times CD)}{2}$

$A=\frac{(2b\sin \theta \times (a-a\cos \theta \left)\right)}{2}$

$A=\frac{2a\left(b\sin \theta \right(1-\cos \theta \left)\right)}{2}$

$A=ab\sin \theta (1-\cos \theta )-\cdot \left(i\right)$

diff with respect to $\theta$,

$\frac{dA}{d\theta }=ab\left[\frac{d}{d\theta }\sin \theta \right(1-\cos \theta \left)\right]$

$\frac{dA}{d\theta }=ab\left[\sin \theta \frac{d}{d\theta }\right(1-\cos \theta )+(1-\cos \theta \left)\frac{d}{d\theta }\sin \theta \right]$

$\frac{dA}{d{\theta }^2}ab[{\sin }^2\theta +\cos \theta -{\cos }^2\theta ]-\cdot \left(ii\right)$

$\frac{da}{dc}=0$

$ab({\sin }^2\theta +\cos \theta -{\cos }^2\theta )=0$

$\cos \theta -\cos 2\theta =0(\therefore {\cos }^2\theta -{\sin }^2\theta =\mathrm{con}2\theta )$

$\cos \theta =\cos 2\theta$

$2\theta =2n\pi \pm \theta$

$\theta =n\pi +\frac{\theta }{2}$

$\theta =n\pi -\frac{\theta }{2}$ where $n=0,\pm 1,\pm 2\dots .$

$\theta =\frac{2\pi }{3}\in (0,\pi )\text{again differetiating eq.}\left(ii\right)$

$\frac{d^{2}A}{d{\theta }^2}=ab[2\sin \theta \cos \theta -\sin \theta +2\sin \theta \sin \theta )$

$\frac{d^{2}A}{d{\theta }^2}=ab[4\sin \theta \cos \theta -\sin \theta ]$

$\left[\frac{d^{2}A}{d{\theta }^2}\right]\theta =2\pi /3^{\prime }\angle \theta$

So, $\theta =2\pi /3,A$ is maximum.

Now, $\theta =2\pi /3$ in the eq (i)

$A=$ ab $\sin \left(\frac{2\pi }{3}\right)\times (1-\cos \left(\frac{2\pi }{3}\right))$

$A=ab\left(\frac{\sqrt{3}}{2}\right)(1+\frac{1}{2})$

$A=\left[\frac{3\sqrt{3}}{4}\right]ab$

Correct option is $\left(C\right)$