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)

The given ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

Let $A\equiv (a\cos \theta ,-b\sin \theta )$
Then, $C\equiv (a\cos \theta ,-b\sin \theta )$
$\Delta =$ Area of $△ABC=\frac{1}{2}\times AC\times BD=AD\times BD$
$=b\sin \theta (a-a\cos \theta )$
$=\frac{1}{2}ab(2\sin \theta -\sin 2\theta )$
Now, $\frac{d\Delta }{d\theta }=\frac{1}{2}ab(2\cos \theta -2\cos 2\theta )=\mathrm{0...}$(1) $\Rightarrow \cos 2\theta =\cos \theta \Rightarrow 2{\cos }^2\theta -\cos \theta -1=0$ $\Rightarrow (2\cos \theta +1)(\cos \theta -1)=0$
$\Rightarrow \cos \theta =\frac{-1}{2}$ or $\cos \theta =1$
If $\theta =0,\Delta =0$, which is not possible.
$\therefore \theta =2\pi /3.$
$\therefore {\Delta }_{\max }=\frac{3\sqrt{3}}{4}ab$ [substitiuting the value of $\theta$ in (1)]