Question

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

Answer 1

Let `(x,y)` be a point on the ellipse `4x^2+y^2=4`.

`Leftrightarrow y^2=4-4x^2 Leftrightarrow y=pm2sqrt{1-x^2}`

The distance `d(x)` between `(x,y)` and `(1,0)` can be expressed as

`d(x)=sqrt{(x-1)^2+y^2}`

by `y^2=4-4x^2`,

`=sqrt{(x-1)^2+4-4x^2}`

by multiplying out

`=sqrt{-3x^2-2x+5}`

Let us maximize `f(x)=-3x^2-2x+5`

`f'(x)=-6x-2=0 Rightarrow x=-1/3` (the only critical value)

`f''(x)=-6 Rightarrow x=-1/3` maximizes `f(x)` and `d(x)`

Since `y=pm2sqrt{1-(-1/3)^2}=pm{4sqrt{2}}/3`,

the farthest points are `(-1/3,pm{4sqrt{2}}/3)`.