Question

Write an equation of the ellipse with foci at (0, +2) and co-vertices at (+1,0

Answer 1

Explanation:

We are given the foci of the ellipse as (0, 2) and (0, -2). "c" is the distance from the center to one focal point. Our center is directly between the 2 foci, so the center is located at (0, 0). "c" then is 2.

We are also given the co-vertex of (1, 0) and (-1, 0) which is located on the minor axis (shorter axis). The foci ALWAYS LIE ON THE MAJOR AXIS, and if we just determined that the shorter axis is the x axis (because the points (1, 0) and (-1, 0) lie on the x-axis) then our ellipse is vertically stretched. The equation for a vertically stretched ellipse is

`(x^2)/(b^2) +(y^2)/(a^2) =1`

As far as the "a" and the "b" go, a is ALWAYS LARGER THAN B, AND WILL ALWAYS LIE UNDER THE AXIS THAT IS THE MAJOR AXIS. Y is our major axis, therefore, a lies under the y-squared fraction. (Also, in an ellipse, the x-squared and the y-squared do not move...only the a and the b do. In a hyperbola, the a and b remain fixed and the x-squared and y-squared move.)

So we know that c = 2 and b = 1, now we need to find a. The equation for the foci of an ellipse is

`c^2=a^2-b^2`

Solving this for a:

`a^2=c^2+b^2` so

`a^2=2^2+1^2` and

`a^2=5`

Filling in our equation now with a and b:

`(x^2)/(1)+(y^2)/(5)=1`