Please explain how it is done. Don't send links .

Solutions:

Ellipse, the area of triangle ABC (A) is given by. 

$$\begin{gathered} A=\frac{1}{2}\overline{)a\left(\frac{2b}{a} \sqrt{a^2-x_1^2}\right)+\left(-x_1\right)}\left(-\frac{b}{a}\sqrt{a^2-x_1^2}\right)+\left(-x_1\right) \\ \Rightarrow A=b\sqrt{a^2-x_1^2} + x_1\frac{b}{a}\sqrt{a^2-x_1^2} \\ \therefore \frac{dA}{d{x}_1}=\frac{-2x_{1}b}{2\sqrt{a^2-x_1^2}}+\frac{\mathbf{b}}{\mathbf{a}}\sqrt{a^2-x_1^2}-\frac{2b{x}_1^2}{a2\sqrt{a^2-x_1^2}} \\ = \frac{b}{a\sqrt{a^2-x_1^2}}\left[-x_{1}a+\left(a^2-x_1^2\right)-x_1^2\right] \\ Now. \frac{dA}{d{x}_1}=0 \\ \Rightarrow -2x_1^2-x_{1}a+a^2=+ \\ \Rightarrow x_1=\frac{a\pm \sqrt{a^2-4\left(-1\right)\left(a^2\right)}}{2\left(-2\right)} \end{gathered}$$