#### Answer

$(-\frac{1}{3},\frac{4\sqrt{2}}{3})~~$ and $~~(-\frac{1}{3},-\frac{4\sqrt{2}}{3})~~$ are the points on the ellipse that are farthest from the point $(1,0)$

#### Work Step by Step

$4x^2+y^2 = 4$
$y^2 = 4-4x^2$
$y = \pm~\sqrt{4-4x^2}$
Note that $-1 \leq x \leq 1$ and $-2 \leq y \leq 2$
We can write an expression for the distance from points on the ellipse to the point $(1,0)$:
$d = \sqrt{(x-1)^2+(y-0)^2}$
$d = \sqrt{(x-1)^2+(y)^2}$
$d = \sqrt{x^2-2x+1+4-4x^2}$
$d = \sqrt{-3x^2-2x+5}$
We can find the point where $d'(x) = 0$:
$d(x) = \sqrt{-3x^2-2x+5}$
$d'(x) = \frac{-6x-2}{2 \sqrt{-3x^2-2x+5}} = 0$
$-6x-2 = 0$
$x = -\frac{1}{3}$
When $0 \leq x \lt -\frac{1}{3}$, then $d'(x) \gt 0$
When $x \gt -\frac{1}{3}$, then $d'(x) \lt 0$
Thus, $x=-\frac{1}{3}$ is the point where the distance $d(x)$ is a maximum.
We can find $y$:
$y = \pm~\sqrt{4-4x^2}$
$y = \pm~\sqrt{4-4(-\frac{1}{3})^2}$
$y = \pm~\sqrt{\frac{32}{9}}$
$y = \pm~\frac{\sqrt{32}}{3}$
$y = \pm~\frac{4\sqrt{2}}{3}$
$(-\frac{1}{3},\frac{4\sqrt{2}}{3})~~$ and $~~(-\frac{1}{3},-\frac{4\sqrt{2}}{3})~~$ are the points on the ellipse that are farthest from the point $(1,0)$