Isosceles triangle inscribed in an ellipse. 
Find the maximum area of an isosceles triangle, which is inscribed inside an ellipse $\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$ with its (unique) vertex lying at one of the ends of the major axis of the ellipse.

The three vertices of the triangle would be $(a,0), (x,y), (x, -y)$.
The area of the triangle by Heron's formula is $$A^2 = (x-a)^2y^2 = (x-a)^2b^2\left( 1-
 \dfrac{x^2}{a^2}\right) \tag{1}.$$
Hence $$\dfrac{dA}{dx} = 0 \implies (x-a)^2 \left( x + \dfrac{a}2\right) = 0.$$
We have minimum at $x = a$ and maximum at $x = -\frac{a}{2}$.
Substituting back in  $(1)$ and taking square roots on both the sides gives $$A = \dfrac{\sqrt{3}ab}{4}$$.
The given answer is $3A$. 
What mistake I made ? Is it possible to find the answer to this question if the triangle was scalene ?

By AM-GM 
$$S_{\Delta}=(a-x)y=(a-x)b\sqrt{1-\frac{x^2}{a^2}}=ab\left(1-\frac{x}{a}\right)\sqrt{1-\frac{x^2}{a^2}}=$$
$$=\frac{ab}{\sqrt3}\cdot\sqrt{\left(1-\frac{x}{a}\right)^3\left(3+\frac{3x}{a}\right)}\leq\frac{ab}{\sqrt3}\cdot\sqrt{\left(\frac{3\left(1-\frac{x}{a}\right)+3+\frac{3x}{a}}{4}\right)^4}=\frac{3\sqrt3ab}{4}.$$
The equality occurs for $1-\frac{x}{a}=3+\frac{3x}{a}$, which says that the equality indeed occurs, 
which says that $\frac{3\sqrt3ab}{4}$ is a maximal value.
Done!
