# Find the maximum area of the isosceles triangle inscribed in the ellipse $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1 with its vertex at one end of major axis. ## 1 Answer Toolbox: • Area of a triangle =\large\frac{1}{2}$$\times l\times h$
• $\large\frac{d}{dx}$$(\cos\theta)=-\sin\theta • \large\frac{d}{dx}$$(\sin\theta)=\cos\theta$
$A$=Area of isosceles $\Delta APP'$
$\;\;=\large\frac{1}{2}$$\times PP'\times AM \;\;=\large\frac{1}{2}$$ab(2b\sin \theta)(a-a\cos\theta)$
$\;\;=ab[\sin\theta-\large\frac{1}{2}$$\sin 2\theta] Step 2: Differentiating with respect to \theta \large\frac{dA}{d\theta}$$=ab(\cos\theta-\large\frac{1}{2}$$\cos 2\theta.2) \quad\;=ab[\cos\theta-\cos 2\theta] For maxima and minima \large\frac{dA}{d\theta}$$=0$
$ab[\cos\theta-\cos 2\theta]=0$
$\cos 2\theta=\cos\theta$
$\theta=\large\frac{2\pi}{3}$
Step 3:
Now,$\large\frac{d^2A}{d\theta}$$=ab[-\sin\large\frac{2\pi}{3}$$+2\sin\large\frac{4\pi}{3}]$
$\qquad\quad\;\;=ab[-\big(\large\frac{\sqrt 3}{2}\big)$$+2\big(\large\frac{-\sqrt 3}{2}\big)] \qquad\quad\;\;=ab[-\big(\large\frac{\sqrt 3}{2}\big)$$-\big(2\large\frac{\sqrt 3}{2}\big)]$
$\qquad\quad\;\;=ab\big[\large\frac{-3\sqrt{3}}{2}\big]$$<0 Step 4: A is maximum when \theta=\large\frac{2\pi}{3}$$=120^\circ$
Maximum value of $A=ab(\sin 120^\circ-$$\large\frac{1}{2}$$\sin 240^\circ)$
$\qquad\qquad\qquad\quad\;\;=ab[\large\frac{\sqrt 3}{2}-\frac{1}{2}\big(\large\frac{-\sqrt 3}{2}\big)]$