Question

Find 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.

Answer 1

Solution:

Consider the isosceles triangle $ABC:$

$A(a,0),B(a\cos \theta ,b\sin \theta )$ and $C(a\cos \theta ,-b\sin \theta )$

Area of $△ABC=\frac{1}{2}\times BC\times$ height of $△ABC$

Height of $△ABC=a(1+\cos \theta )$

$BC=2b\sin \theta$

or, $△=\frac{1}{2}\times 2b\sin \theta \times a(1+\cos \theta )=ab\sin \theta (1+\cos \theta )$

For maximum area of the triangle,

$\frac{d△}{d\theta }=ab\cos \theta (1+\cos \theta )-absi{n}^2\theta =0$

or, $\cos \theta (1+\cos \theta )-si{n}^2\theta =0$

or, $\cos \theta (1+\cos \theta )-(1+\cos \theta )(1-\cos \theta )=0$

or, $(1+\cos \theta )(\cos \theta -1+\cos \theta )=0$

or, $(1+\cos \theta )(2\cos \theta -1)=0$

or, $\cos \theta =-1$ or, $\cos \theta =\frac{1}{2}$

For maximum value, we take

$\cos \theta =\frac{1}{2}$ and $\sin \theta =\frac{\sqrt{3}}{2}$

or, Maximum area $=ab\times \frac{\sqrt{3}}{2}(1+\frac{1}{2})=\frac{3\sqrt{3}}{4}ab$