Let the maximum area of the triangle that can be inscribed in the ellipse x^2a^2+y^24=1, a…

Question

Let the maximum area of the triangle that can be inscribed in the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}4=1$, $a>2$, having one vertex at one end of the major axis and one side parallel to the $y$-axis, be $6\sqrt3$. Then the eccentricity of the ellipse is:

Accepted Answer

With the fixed vertex at $(a,0)$ and the base at $x=a\cos\theta$: Area $=2a\sin\theta(1-\cos\theta)$ (using $b=2$). Maximizing over $\theta$: $2\cos^2\theta-\cos\theta-1=0\Rightarrow\cos\theta=-\tfrac12$ (rejecting $\cos\theta=1$).

At $\theta=120^\circ$: Area$_{\max}=2a\cdot\dfrac{\sqrt3}2\cdot\dfrac32=\dfrac{3\sqrt3}2a$.

$$\frac{3\sqrt3}2a=6\sqrt3\ \Longrightarrow\ a=4,\ a^2=16,\ b^2=4.$$

$$e=\sqrt{1-\frac4{16}}=\frac{\sqrt3}2.$$

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