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 ?

Everything is correct until you use the value of $x$ to calculate the area.
You should have $$A^2=(-\frac{3a}{2})^2b^2(1-\frac 14)$$ which will give you the correct answer $$A=\frac{3\sqrt{3}}{4}ab$$
