Answer:
Step-by-step 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
[tex]\frac{x^2}{b^2} +\frac{y^2}{a^2} =1[/tex]
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
[tex]c^2=a^2-b^2[/tex]
Solving this for a:
[tex]a^2=c^2+b^2[/tex] so
[tex]a^2=2^2+1^2[/tex] and
[tex]a^2=5[/tex]
Filling in our equation now with a and b:
[tex]\frac{x^2}{1}+\frac{y^2}{5}=1[/tex]
