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. $None\, of\, these$

Answer 1

Answer: (B)

$\frac{3\sqrt{3}}{4} ab$

Answer 2

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
$A = \frac{1}{2} \times AD \times BC$
$= \frac{1}{2} \times\left(a+x\right)\times2y =y\left(a+x\right)$
$=\sqrt{b^{2}\left(1-\frac{x^{2}}{a^{2}}\right)} \left(a+x\right) =\frac{b}{a} \sqrt{a^{2} -x^{2}} \left(a+x\right)$
$\Rightarrow \frac{dA}{dx} =\frac{b}{a} \left[\left(a+x\right) . \frac{1}{2} \times\frac{-2x}{\sqrt{a^{2}-x^{2}}} +\sqrt{a^{2} -x^{2}}\right]$
$=\frac{b}{a} \left[\frac{-x\left(a+x\right)+\left(a^{2} -x^{2}\right)}{\sqrt{a^{2}-x^{2}}} \right]$
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 $\frac{d^2A}{dx^2} < 0$ for $x = \frac{a}{2}$
$\therefore$ Max. Area $= \frac{b}{a} . \left( a + \frac{a}{2} \right) \sqrt{a^2 - \frac{a^2}{4}}$
$=\frac{b}{a} . \frac{3}{2}a. \frac{\sqrt{3}a}{2} =\frac{3\sqrt{3} ab}{4}$ sunits