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

• A

  $\sqrt{3} a b$ sq units
• Correct answer
  B

  $\frac{3 \sqrt{3}}{4} a b$ sq units
• C

  $\frac{5 \sqrt{3}}{4} a b$ sq units
• D

  None of these

Solution

Let $\triangle A B C$ be an isosceles triangle inscribed in the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. Let coordinate of $A$ be $(a, 0)$. Area of $\triangle A B C, A=\frac{1}{2} \times A D \times B C$ $$ \begin{aligned} &=\frac{1}{2} \times(a+x) \times 2 y=y(a+x) \\ &=\sqrt{b^{2}\left(1-\frac{x^{2}}{a^{2}}\right)}(a+x)=\frac{b}{a} \sqrt{a^{2}-x^{2}}(a+x) \end{aligned} $$

$\Rightarrow \frac{d A}{d x}=\frac{b}{a}\left[(a+x) \frac{1}{2} \times \frac{-2 x}{\sqrt{a^{2}-x^{2}}}+\sqrt{a^{2}-x^{2}}\right]$ $=\frac{b}{a}\left[\frac{-x(a+x)+\left(a^{2}-x^{2}\right)}{\sqrt{a^{2}-x^{2}}}\right]$ Put $\frac{d A}{d x}=0$ $\Rightarrow \quad 2 x^{2}+a x-a^{2}=0$ $\Rightarrow \quad 2 x^{2}+2 a x-a x-a^{2}=0$ $\Rightarrow \quad(2 x-a)(x+a)=0$ $\Rightarrow \quad x=\frac{a}{2},-a$ Since, $\frac{d^{2} A}{d x^{2}} < 0$, for $x=\frac{a}{2}$ $$ \begin{aligned} \therefore \text { Maximum area } &=\frac{b}{a} \cdot\left(a+\frac{a}{2}\right) \sqrt{a^{2}-\frac{a^{2}}{4}} \\ &=\frac{b}{a} \cdot \frac{3}{2}(a) \frac{\sqrt{3} a}{2}=\frac{3 \sqrt{3} a b}{4} \text { sq units } \end{aligned} $$

Hence, the correct answer is option: 'B'.