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

Jun 27, 2017

Explanation:

$4 {y}^{2} - 9 {x}^{2} = 36$ 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 ${\left(y - k\right)}^{2} / {a}^{2} + {\left(x - h\right)}^{2} / {b}^{2} = 1$

where $\left(h , k\right)$ is the center and vertices are $\left(h , k + a\right)$ and $\left(h , k - a\right)$ and equation of asymptotes are $y = \pm \frac{a}{b} \left(x - h\right) + k$.

as $4 {y}^{2} - 9 {x}^{2} = 36$ can be written as ${\left(y - 0\right)}^{2} / {3}^{2} - {\left(x - 0\right)}^{2} / {2}^{2} = 1$ and as such

center is $\left(0 , 0\right)$ and vertices are $\left(0 , 3\right)$ and $\left(0 , - 3\right)$, while asymtotes are $3 x - 2 y = 0$ and $3 x + 2 y = 0$.

The graph appears as shown below.

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