Question

The maximum area of an isosceles triangle inscribed in the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with its vertex at one end of the major axis is:

• A. $\sqrt{3}ab$
• B. $\frac{3\sqrt{3}}{4}ab$
• C. $\frac{5\sqrt{3}}{4}ab$
• D. None of these

Answer 1

Answer: (B)

$\frac{3\sqrt{3}}{4}ab$

In $△ABC$,

Base $=2a\cos \theta$

Height $=b-(-b\sin \theta )=b(1+\sin \theta )$

$\therefore$ Area $E=\frac{1}{2}\left(2a\cos \theta \right)\left(b\right(1+sin\theta \left)\right)$ -----(1)

For maximum area, $\frac{dE}{d\theta }=0$

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

$\Rightarrow \cos \left(2\theta \right)=\cos (\frac{\pi }{2}-\theta )$

$\Rightarrow \theta =\frac{\pi }{6}$

Putting the value in equation (1) we get:

$E=\frac{1}{2}\left(2a\frac{\sqrt{3}}{2}\right)\left(b\right(1+\frac{1}{2}\left)\right)$

$E=\frac{3\sqrt{3}}{4}ab$