Find the points on the ellipse 3x2 + y2 = 3 that are farthest away from the point (−1, 0).

Question

Find the points on the ellipse 3x² + y ² = 3 that is farthest away from the point (−1, 0).

Answer 1

Just to make the math a little easier, consider the points on the ellipse to be (cos theta, √3 sin theta)

Then, we minimize their distance squared from (-1, 0) - incidentally, a point on the ellipse, is

(cos theta - -1)^2 + (√3 sin theta)^2 =

cos^2 theta + 2 cos theta + 1 + 3 sin^2 theta = (substituting sin^2 theta + cos^2 theta = 1)

2 + 2 cos theta + 2 sin^2 theta

The derivative is

-2 sin theta + 4 sin theta cos theta =

-2 sin theta (2 cos theta - 1)

This equals 0 at sin theta = 0 or cos theta = 1/2, which gives us values of 0°, 180°, 60°, 300°

If we rewrite the first derivative as - 2 sin theta + 2 sin 2 theta, the second derivative is

-2 cos theta + 4 cos 2 theta

At 0°, 180°, 60°, 300°, we get

2, 6, -3, -3

Thus, we have relative minima at 0°, 180° and relative maxima at 60°, 300°

Clearly, given the nature of the function, the global minima and maxima will be among these four locations.

Also, clearly, 180° is the global minimum, as it is at the actual point. 60° and 300° gives the same value of the distance squared, so they will each be maxima.

(cos theta, √3 sin theta) at 60°, 300° is (1/2, 3/2) and (1/2, -3/2)