The points on the ellipse 4x^2+y^2=4 that are farthest from the point…

The correct answer is Option (3) → \( \left(\displaystyle\frac{-1}{3},\pm\displaystyle\frac{4\sqrt{2}}{3}\right) \)

Let (x, y) be a point on the ellipse. It is easier to maximise the square of the distance from (1, 0). On the ellipse

    \( \displaystyle y^{2}=4-4x^{2},\quad -1\le x\le1 \)

so

    \( \displaystyle D^{2}=(x-1)^{2}+y^{2} \)

    \( \displaystyle =x^{2}-2x+1+4-4x^{2} \)

    \( \displaystyle =-3x^{2}-2x+5 \)

This is a downward parabola in x, largest at its vertex:

    \( \displaystyle x=-\frac{b}{2a}=-\frac{-2}{2(-3)} \)

    \( \displaystyle =-\frac{1}{3} \)

which lies in \( [-1,1] \). Then

    \( \displaystyle y^{2}=4-\frac{4}{9} \)

    \( \displaystyle =\frac{32}{9} \)

    \( \displaystyle y=\pm\frac{4\sqrt{2}}{3} \)

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