Question

Find an equation in standard form for the hyperbola with vertices at (0, ±6) and foci at (0, ±9).

Answer 1

`(y^2)/(36)-(x^2)/(45)=1.`

Explanation:

Since vertices and foci lie on the y-axis, the equation of the hyperbola is

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

If the vertices are at points (0,±6), then
`b=6.`

If the foci are at points (0,±9), then
`c=9.`

Note that

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

then

`9^2=6^2+a^2,\\ \\a^2=81-36,\\ \\a^2=45.`

The equation of the hyperbola is

`(y^2)/(36)-(x^2)/(45)=1.`