Elliptical Distances

Let $(x, y)$ be a point on the ellipse. The distance between $(x, y)$ and the given point $(-1, 0)$ is given by $\sqrt{(x + 1)^2 + y^2}$

Using the fact that $4x^2 + y^2 = 4 \rightarrow y^2 = 4 - 4x^2$ , we rewrite the distance above as $\sqrt{(x + 1)^2 + (4 - 4x^2)} \rightarrow \sqrt{-3x^2 + 2x + 5}$

This quantity is maximized when the quadratic inside the radical uses $x = -\frac{b}{2a} = \frac{1}{3}$ .

To be a maximum, the tangent at $T$ must be perpendicular to $PT$ .

Slope of the tangent at point $(x;y)$ : $(4x^2+y^2)'=4'~\Rightarrow~8x+2yy'=0~\Leftrightarrow~y'=\frac{-4x}{y}$ . At $T$ , it is $\frac{-4a}{b}$ .

Slope of $PT$ is $\frac{b}{a+1}$ .

The lines are perpendicular $\Leftrightarrow$ the product of their slope is $-1$ : $\frac{-4a}{b}·\frac{b}{a+1}=-1~\Rightarrow~4a=a+1~\Leftrightarrow~\boxed{a=\frac{1}{3}}$ .