Question

find the equation for the graph of the hyperbola with vertices at (0, 4) and (0, -4) and one focus at (0, -9).

Explanation:

Step1: Determine the form of the hyperbola

The vertices are at $(0,4)$ and $(0, - 4)$, so the hyperbola has a vertical transverse - axis. The standard form of a hyperbola with a vertical transverse axis is $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}} = 1$. The center of the hyperbola is the mid - point of the vertices. The mid - point of $(0,4)$ and $(0, - 4)$ is $(0,0)$. The distance from the center $(0,0)$ to a vertex is $a$. So, $a = 4$.

Step2: Find the value of $c$

The focus is at $(0,-9)$. The distance from the center $(0,0)$ to a focus is $c$. So, $c = 9$.

Step3: Calculate the value of $b^{2}$

Use the relationship $c^{2}=a^{2}+b^{2}$. Substitute $a = 4$ and $c = 9$ into the equation: $9^{2}=4^{2}+b^{2}$. Then $b^{2}=c^{2}-a^{2}=81 - 16=65$.

Step4: Write the equation of the hyperbola

Substitute $a^{2}=16$ and $b^{2}=65$ into the standard form $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}} = 1$. The equation of the hyperbola is $\frac{y^{2}}{16}-\frac{x^{2}}{65}=1$.

Answer:

$\frac{y^{2}}{16}-\frac{x^{2}}{65}=1$