Let the maximum area of the triangle that can be inscribed in the ellipse [x^2 a^2 + y^2 4…

Question

Let the maximum area of the triangle that can be inscribed in the ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over 4} = 1,\,a > 2$$, having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the y-axis, be $$6\sqrt 3 $$. Then the eccentricity of the ellipse is :

Accepted Answer

Given ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over 4} = 1,\,a > 2$$

$$\therefore$$ Let A($$\theta$$) be the area of $$\Delta$$ABB'

Then $$A(\theta ) = {1 \over 2}4\sin \theta (a + a\cos \theta )$$

$$A'(\theta ) = a(2\cos \theta  + 2{\cos ^2}\theta )$$

For maxima $$A'(\theta ) = 0$$

$$ \Rightarrow \cos \theta  = 1,\,\,\cos \theta  = {1 \over 2}$$

But for maximum area $$\cos \theta  = {1 \over 2}$$

$$\therefore$$ $$A(\theta ) = 6\sqrt 3 $$

$$ \Rightarrow 2{{\sqrt 3 } \over 2}\left( {a + {a \over 2}} \right) = 6\sqrt 3 $$

$$ \Rightarrow a = 4$$

$$\therefore$$ $$e = \sqrt {1 - {{{b^2}} \over {{a^2}}}}  = \sqrt {1 - {4 \over {16}}}  = {{\sqrt 3 } \over 2}$$

Solutions · 1