Question

Find the area of the greatest of isosceles triangle that can be inscribed in a given ellipse having its vertex coincide with one end of major axes?

Answer 1

to find the area of the greatest of isosceles triangle that which can be inscribed in to a given ellipse having its vertex coincide with one end of major axes.

as we know that ellipse equation is $\frac{x^2}{a^2}+\frac{a^2}{b^2}=1$

Let us find out the area of $△AP\theta$

as we know that area of triangle=$\frac{1}{2}\times b\times n$

$\Rightarrow △APQ=\frac{1}{2}PQ.AD$

$\frac{1}{2}.2\left(PD\right)(OA-OD)$ $fig\left(a\right)$ Assumption of the ellipse as according to given data.

$\left(b\sin \theta \right)[a-a\cos \theta ]\_\_\_\_\_\_eq\left(1\right)$ $p=[a\cos \theta ,b\sin \theta ]\because$ it is a vertex of an ellipse.

$eq\left(1\right)[\because$ from the figure configuration$]$ $Q=[a\cos \theta ,-b\sin \theta ]\because$ it is perpendicular to the ellipse

$\Rightarrow ab\sin \theta -ab\sin \theta \cos \theta \iff ab[\sin \theta -\sin \theta \cos \theta ]$

Now , by multiplying and dividing by $2$ we get,

$△APQ=\frac{ab}{2}[2\sin \theta -2\sin \theta \cos \theta ]$

$△APQ\Rightarrow \frac{ab}{2}[2\sin \theta -\sin 2\theta ]\_\_\_\_eq\left(2\right)$

by differentiating $eq\left(2\right)$ we get , $\frac{d△}{d\theta }=0$

$\Rightarrow \frac{ab}{2}\left[\frac{d}{d\theta }\right[2\sin \theta ]-\frac{d}{d\theta }[\sin 2\theta \left]\right]$

$\Rightarrow \frac{ab}{2}[2\cos \theta -2\cos 2\theta ]=0$

$\Rightarrow [2\cos \theta -2\cos \theta ]=0$ $[\because \cos 2\theta =2{\cos }^2\theta -1]$

$\Rightarrow 2\cos \theta -2[2{\cos }^2\theta -1]=0$

$\Rightarrow 2[\cos \theta -[2co{s}^2\theta -1\left]\right]=0$

$\Rightarrow \cos \theta -2{\cos }^2\theta +1=0$

$-[2{\cos }^2\theta -\cos \theta -1]=0$

$\Rightarrow 2co{s}^2\theta -\cos \theta -1=0$ by factorization we get,

$2{\cos }^2\theta -2\cos \theta +\cos \theta -1=0$

$2\cos \theta [\cos \theta -1]+1[\cos \theta -1]=0$

$\Rightarrow (2\cos \theta +1)(\cos \theta -1)=0$ (thus factors obtained)

$\Rightarrow cos\theta =\frac{-1}{2};\cos \theta =1$

for $\cos \theta =\frac{-1}{2};\theta =\frac{2\pi }{3}\theta >0$ this can be taken.

for $\cos \theta =1;\theta =0\theta \ne 0$this condition cant be taken

So, we get

$\Rightarrow \cos \theta =\frac{-1}{2}\Rightarrow \theta =\frac{2\pi }{3}$ $[\because$ maximum value of $\theta$ is $\frac{2\pi }{3}]$

Now , let us find the Area for $\frac{ab}{2}[2\sin \theta -2\sin \theta \cos \theta ]$

for $\theta =\frac{2\pi }{3}$

Area $=\frac{ab}{2}\left[2\sin \left[\frac{2\pi }{3}\right]\right]-\left[2\sin \frac{2\pi }{3}\right]\cos \left[\frac{2\pi }{3}\right]$

$\Rightarrow \frac{ab}{2}\times 2[\sin \left[\frac{2\pi }{3}\right]-\sin \left[\frac{2\pi }{3}\right]\cos \left[\frac{2\pi }{3}\right]]$

$\Rightarrow ab[\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}\left[\frac{-1}{2}\right]]\Rightarrow ab[\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{4}]$

$\Rightarrow ab\left[\frac{2\sqrt{3}+\sqrt{3}}{4}\right]\Rightarrow ab\left[\frac{2\sqrt{3}}{4}\right]$

$\therefore Area=\frac{3\sqrt{3}ab}{4}$ is the maximum concidence of major axis.