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

Question

95880 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-09-19T19:19:20

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)#.