Question

5. [Type C] (10 pts) Find the point(s) on the ellipse \( 4 x^{2}+y^{2}=4 \) that are the farthest from the point \( (1,0) \).

Answer 1

`(-(1)/(3),(4√(2))/(3))` and
`(-(1)/(3),-(4√(2))/(3))`

Explanation:

We know that the distance between some point
`(x,y)` and
`(1,0)` is
`d=√((x-1)^2+y^2)` where
`y^2=4-4x^2` from the equation of the ellipse:

`d(x)=√((x-1)^2+(4-4x^2))\\d(x)=√(x^2-2x+1+4-4x^2)\\d(x)=√(-3x^2-2x+5)`

Distance is maximized where
`d'(x)=0`, and whatever that critical value is, we can get our x and y-coordinates of the points farthest from (1,0):

`d'(x)=(-6x-2)/(2√(-3x^2-2x+5))\\\\0=(-6x-2)/(2√(-3x^2-2x+5))\\\\0=-6x-2\\\\2=-6x\\\\x=-(1)/(3)`

Now we can get the y-coordinate(s) of the point farthest from (1,0):

`y^2=4-4x^2\\y^2=4-4(-(1)/(3))^2\\y^2=4-4((1)/(9))\\y^2=4-(4)/(9)\\y^2=(32)/(9)\\y=\pm(√(32))/(3)\\y=\pm(4√(2))/(3)`

Therefore, the points on the ellipse that are the farthest from the point (1,0) are
`(-(1)/(3),(4√(2))/(3))` and
`(-(1)/(3),-(4√(2))/(3))`