# How do you find the foci and sketch the hyperbola 9y^2-x^2=4?

Dec 27, 2016

#### Explanation:

The standard form for a hyperbola of this type is:

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

where, x and y correspond to any point $\left(x , y\right)$ on the hyperbola, h and k correspond to the center point, $\left(h , k\right)$, "a" is the vertical distance from the center to the vertices, and $\sqrt{{a}^{2} + {b}^{2}}$ is the vertical distance from the center to the foci.

Given: $9 {y}^{2} - {x}^{2} = 4$

We will be using algebraic steps to put the above equation into the standard form.

Divide both sides by 4 in the form of ${2}^{2}$:

$9 {y}^{2} / {2}^{2} - {x}^{2} / {2}^{2} = 1$

Write the 9 as ${3}^{2}$ and the denominator of the divisor:

${y}^{2} / \left({2}^{2} / {3}^{2}\right) - {x}^{2} / {2}^{2} = 1$

Group the denominator of the first term into a single fraction:

${y}^{2} / \left({\left(\frac{2}{3}\right)}^{2}\right) - {x}^{2} / {2}^{2} = 1$

$\sqrt{{a}^{2} + {b}^{2}} = \sqrt{{\left(\frac{2}{3}\right)}^{2} + {2}^{2}} = \frac{2 \sqrt{10}}{3}$

The foci are located at $\left(0 , - \frac{2 \sqrt{10}}{3}\right) \mathmr{and} \left(0 , \frac{2 \sqrt{10}}{3}\right)$

To help you sketch the hyperbola, the center is $\left(0 , 0\right)$ and the vertices are at $\left(0 , - \frac{2}{3}\right) \mathmr{and} \left(0 , \frac{2}{3}\right)$

There is a graph of the original equation: 