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Q.
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
Conic Sections
Solution:
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 $\triangle A B C=\frac{1}{2} \times A C \times B D=A D \times B D$
$=b \sin \theta(a-a \cos \theta)$
$=\frac{1}{2} a b(2 \sin \theta-\sin 2 \theta)$
Now, $\frac{d \Delta}{d \theta}=\frac{1}{2} a b(2 \cos \theta-2 \cos 2 \theta)=0 ...$(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.
$\therefore \theta=2 \pi / 3 .$
$\therefore \Delta_{\max }=\frac{3 \sqrt{3}}{4} a b$
[substitiuting the value of $\theta$ in (1)]
