Question

How do you find the center and vertices and sketch the hyperbola `4y^2-9x^2=36`?

Answer 1

Please see below.

Explanation:

`4y^2-9x^2=36` is the equation of a vertical hyperbola, wherein center, foci and vertices line up above and below each other, parallel to the `y`-axis.

The general form of such an equation is `(y-k)^2/a^2+(x-h)^2/b^2=1`

where `(h,k)` is the center and vertices are `(h,k+a)` and `(h,k-a)` and equation of asymptotes are `y=+-a/b(x-h)+k`.

as `4y^2-9x^2=36` can be written as `(y-0)^2/3^2-(x-0)^2/2^2=1` and as such

center is `(0,0)` and vertices are `(0,3)` and `(0,-3)`, while asymtotes are `3x-2y=0` and `3x+2y=0`.

The graph appears as shown below.

graph{(4y^2-9x^2-36)(3x-2y)(3x+2y)=0 [-20, 20, -10, 10]}