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

1 Answer
Sep 19, 2014

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).