Question

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

Answer 1

The given ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Let the major axis be along the x−axis.
Let ABC be the triangle inscribed in the ellipse where vertex C is at (a,0).
Since the ellipse is symmetrical with respect to the x−axis and y−axis, we can assume the coordinates of A to be $(−x_1,y_1)$ and the coordinates of B to be $(−x_1,−y_1)$.
Now, we have $y_1=±\frac{b}{a}\sqrt{a^2-x^2_1}$
∴Coordinates of A are $(-x_1,\frac{b}{a}\sqrt{a^2-x^2_1})$ and the coordinates of B are
$(-x_1,\frac{b}{a}\sqrt{a^2-x^2_1})$
As the point $(x_1,y_1)$ lies on the ellipse, the area of triangle ABC(A) is given by,
$A=\frac{1}{2}|a(\frac{2b}{a}\sqrt{a^2-x^2_1})+(-x_1)$ $(\frac{b}{a}\sqrt{a^2-x^2_1})+(-x_1)(\frac{b}{a}\sqrt{a^2-x^2_1})+(-x_1)$
$⇒A=b\sqrt{a^2-x_1^2}+x_1\sqrt{a^2-x_1^2} ...(1)$
∴\frac{dA}{dx_1}$$=-\frac{2x_1b}{2\sqrt{a^2-x^2_1}}+\frac{b}{a}{\sqrt{a^2-x^2_1}}-\frac{2bx^2_1}{a2\sqrt{a^2-x^2_1}}
$=\frac{b}{a\sqrt{a^2-x_1^2}}[-x_1a+(a^2-x_1^2)-x_1^2]$
$=\frac{b(-2x_1^2-x_1a+a^2)}{a\sqrt{a^2-x_1^2}}$

Now,$\frac{dA}{dx_1}=0$
$⇒-2x_1^2-x_1a+a^2=0$
$⇒x_1=\frac{a±\sqrt{a^2-4(-2)(a^2)}}{2(-2)}$
$=\frac{a±\sqrt{9a^2}}{-4}$
$=\frac{a±3a}{-4}$
$⇒x_1=-a,\frac{a}{2}$
But, $x_1$ cannot be equal to a.
$∴x_1=\frac{a}{2}⇒y_1=\frac{b}{a}\sqrt{a^2-\frac{a^2}{4}}=\frac{ba}{2a}=\sqrt3=\frac{\sqrt{3b}}{2}$

Now$\frac{d^2A}{dx^2_1}$=\frac{b}{a}$$\bigg[\frac{\sqrt{a^2-x_1^2}(-4x_1-a)-(-2x^2_1-x_1a+a^2)\frac{(-2x_1)}{\sqrt{a^2-x^2_1}}}{a^2-x^2_1}\bigg]

$=\frac{b}{a}[\frac{a^2-x^2_1(-4x_1-a)+x_1-(-2x^2_1-x_1a+a^2)}{(a^2-x_1^2)^{\frac{3}{2}}}]$
$=\frac{b}{a}\frac{2x^3-3a^2x-a^3}{(a^2-x^2_1)^\frac{3}{2}}$

Also, when $x_1=\frac{a}{2}$,,then
$\frac{d^2A}{dx^2_1}=\frac{b}{a}\bigg[\frac{\frac{2a3}{8}-3\frac{a^3}{2}-a^3}{(\frac{3a^2}{4})^\frac{3}{2}}\bigg]$
$=\frac{-b}{a}\bigg[\frac{\frac{9}{4}a^3}{(\frac{3a^2}{2})^{\frac{3}{2}}}\bigg]<0$
Thus, the area is the maximum when $x_1=\frac{a}{2}.$
∴ Maximum area of the triangle is given by,
$A=b\sqrt{a^2-\frac{a^2}{4}}+(\frac{a}{2})\sqrt{a^2-\frac{a^2}{4}}$
$=ab\frac{\sqrt3}{2}+(\frac{a}{2})\frac{b}{a}\times\frac{a\sqrt3}{2}$
$=\frac{ab√3}{2}+\frac{ab√3}{4}=\frac{3√3}{4}ab$