# How do you find the points on the ellipse 4x^2+y^2=4 that are farthest from the point (1,0)?

Sep 19, 2014

Let $\left(x , y\right)$ be a point on the ellipse $4 {x}^{2} + {y}^{2} = 4$.

$\Leftrightarrow {y}^{2} = 4 - 4 {x}^{2} \Leftrightarrow y = \pm 2 \sqrt{1 - {x}^{2}}$

The distance $d \left(x\right)$ between $\left(x , y\right)$ and $\left(1 , 0\right)$ can be expressed as

$d \left(x\right) = \sqrt{{\left(x - 1\right)}^{2} + {y}^{2}}$

by ${y}^{2} = 4 - 4 {x}^{2}$,

$= \sqrt{{\left(x - 1\right)}^{2} + 4 - 4 {x}^{2}}$

by multiplying out

$= \sqrt{- 3 {x}^{2} - 2 x + 5}$

Let us maximize $f \left(x\right) = - 3 {x}^{2} - 2 x + 5$

$f ' \left(x\right) = - 6 x - 2 = 0 R i g h t a r r o w x = - \frac{1}{3}$ (the only critical value)

$f ' ' \left(x\right) = - 6 R i g h t a r r o w x = - \frac{1}{3}$ maximizes $f \left(x\right)$ and $d \left(x\right)$

Since $y = \pm 2 \sqrt{1 - {\left(- \frac{1}{3}\right)}^{2}} = \pm \frac{4 \sqrt{2}}{3}$,

the farthest points are $\left(- \frac{1}{3} , \pm \frac{4 \sqrt{2}}{3}\right)$.