Find the length of the major and minor axes of the ellipse.

Question

4x2+8y2=32

Answer 1

For an ellipse equation in standard form, the length of the major axis is 2a, length of minor is 2b, for

(x-h)²/a² + (y-k)²/b² = 1. (h,k) is the center of the ellipse, which in this case is just (0,0). Note that the a² and b² can change positions in this equation. a is just the bigger of those 2 parameters.

Turning to the specific question, we need to divide both sides of the equation by 32, so we have it = 1:

x²/8 + y²/4 = 1. So a = 2√2 , and b = 2. The ellipse is oriented horizontally (its major axis along the x-axis), with a major axis length = 4√2 and a minor axis length = 4.