Question

The maximum area in square units of an isosceles triangle inscribed in an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with its vertex at one end of the major axis is

• A. $\sqrt{3}ab$
• B. $\frac{3\sqrt{3}}{4}ab$
• C. $\frac{5\sqrt{3}}{4}ab$
• D. $Noneofthese$

Answer 1

Answer: (B)

Let $\Delta ABC$ be an isosceles triangle inscribed in the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Let co-ordinate of $A$ be $(a,0)$
Area of $\Delta ABC$ is
$A=\frac{1}{2}\times AD\times BC$
$=\frac{1}{2}\times \left(a+x\right)\times 2y=y\left(a+x\right)$
$=\sqrt{b^2\left(1-\frac{x^2}{a^2}\right)}\left(a+x\right)=\frac{b}{a}\sqrt{a^2-x^2}\left(a+x\right)$
$\Rightarrow \frac{dA}{dx}=\frac{b}{a}\left[\left(a+x\right).\frac{1}{2}\times \frac{-2x}{\sqrt{a^2-x^2}}+\sqrt{a^2-x^2}\right]$
$=\frac{b}{a}\left[\frac{-x\left(a+x\right)+\left(a^2-x^2\right)}{\sqrt{a^2-x^2}}\right]$
Put $\frac{dA}{ax}=0$
$\Rightarrow 2x^2+ax-a^2=0$
$\Rightarrow 2x^2+2ax-ax-a^2=0$
$\Rightarrow (2x-a)(x+a)=0$
$\Rightarrow x=\frac{a}{2},-a$
Since $\frac{d^{2}A}{d{x}^2}<0$ for $x=\frac{a}{2}$
$\therefore$ Max. Area $=\frac{b}{a}.\left(a+\frac{a}{2}\right)\sqrt{a^2-\frac{a^2}{4}}$
$=\frac{b}{a}.\frac{3}{2}a.\frac{\sqrt{3}a}{2}=\frac{3\sqrt{3}ab}{4}$ sq.units