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,
\(A=b\sqrt{a^2-\frac{a^2}{4}}+(\frac{a}{2})\sqrt{a^2-\frac{a^2}{4}}\)
\(=ab\frac{\sqrt3}{2}+(\frac{a}{2})\frac{b}{a}\times\frac{a\sqrt3}{2}\)
\(=\frac{ab√3}{2}+\frac{ab√3}{4}=\frac{3√3}{4}ab\)