Question

Find the points on the ellipse 4x2 + y2 = 4 that are farthest away from the point (−1, 0).

Answer 1

The two points that are most distant from (-1,0) are exactly (1/3, 4sqrt(2)/3) and (1/3, -4sqrt(2)/3) approximately (0.3333333, 1.885618) and (0.3333333, -1.885618) Rewriting to express Y as a function of X, we get 4x^2 + y^2 = 4 y^2 = 4 - 4x^2 y = +/- sqrt(4 - 4x^2) So that indicates that the range of values for X is -1 to 1. Also the range of values for Y is from -2 to 2. Additionally, the ellipse is centered upon the origin and is symmetrical to both the X and Y axis. So let's just look at the positive Y values and upon finding the maximum distance, simply reflect that point across the X axis. So y = sqrt(4-4x^2) distance is sqrt((x + 1)^2 + sqrt(4-4x^2)^2) =sqrt(x^2 + 2x + 1 + 4 - 4x^2) =sqrt(-3x^2 + 2x + 5) And to simplify things, the maximum distance will also have the maximum squared distance, so square the equation, giving -3x^2 + 2x + 5 Now the maximum will happen where the first derivative is equal to 0, so calculate the first derivative. d = -3x^2 + 2x + 5 d' = -6x + 2 And set d' to 0 and solve for x, so 0 = -6x + 2 -2 = -6x 1/3 = x So the furthest point will be where X = 1/3. Calculate those points using (1) above. y = +/- sqrt(4 - 4x^2) y = +/- sqrt(4 - 4(1/3)^2) y = +/- sqrt(4 - 4(1/9)) y = +/- sqrt(4 - 4/9) y = +/- sqrt(3 5/9) y = +/- sqrt(32)/sqrt(9) y = +/- 4sqrt(2)/3 y is approximately +/- 1.885618