Question

Find the maximum area of an isosceles triangle inscribed in the ellipse with its vertex at one end of the major axis.

Answer 1

The equation of the ellipse is given as,

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

Compare equation (1) with ${\cos }^2\theta +{\sin }^2\theta =1$.

$\frac{x}{a}=\cos \theta$

$\Rightarrow x=a\cos \theta$

$\frac{y}{b}=\sin \theta$

$\Rightarrow y=b\sin \theta$

So, the points on the ellipse is $(a\cos \theta ,b\sin \theta )$.

Consider the diagram shown above and draw a perpendicular $PM$ to $x$-axis up to the point $Q$.

$OM=a\cos \theta$

$PM=b\sin \theta$

Since ellipse is symmetrical about $x$-axis,

therefore $PM=QM$ and hence $\Delta APQ$ is isosceles.

Write the expression of the area of the triangle.

$A=\frac{1}{2}\times PQ\times AM$

$=\frac{1}{2}\times 2PM\times AM$

$=PM(OA-OM)$

Simplify further by substituting the values,

$A=b\sin \theta (a-a\cos \theta )$

$=ab(\sin \theta -\sin \theta \cos \theta )$

$=\frac{ab}{2}(2\sin \theta -2\sin \theta \cos \theta )$

$=\frac{ab}{2}(2\sin \theta -\sin 2\theta )$

Differentiate both sides of the equation,

$\frac{dA}{d\theta }=\frac{ab}{2}(2\cos \theta -2\cos 2\theta )=ab(\cos \theta -\cos 2\theta )$

Again differentiate the both sides of the equation,

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

Now

$\frac{dA}{d\theta }=0$

$\Rightarrow ab(\cos \theta -\cos 2\theta )=0$

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

$\Rightarrow \cos \theta =\cos ({360}^{\circ }-2\theta )$

Simplify further

$\theta ={360}^{\circ }-2\theta$

$\Rightarrow 3\theta ={360}^{\circ }$

$\Rightarrow \theta ={120}^{\circ }$

at $\theta ={120}^{\circ }$,

$\Rightarrow \frac{d^{2}A}{d{\theta }^2}=ab(-\sin {120}^{\circ }+2\sin {240}^{\circ })$

$=ab(\frac{-\sqrt{3}}{2}$ $-\frac{2\sqrt{3}}{2}$ )

$=-ab\left(\frac{3\sqrt{3}}{2}\right)$ <0

A is maximum at $\theta ={120}^{\circ }$.

The expression for the maximum area is,

$A=\frac{ab}{2}(2\sin {120}^{\circ }-\sin {240}^{\circ })$

$=\frac{ab}{2}(\frac{2\sqrt{3}}{2}+\frac{\sqrt{3}}{2})$

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

Thus, the maximum area is $\frac{3\sqrt{3}}{4}ab$.