Question

the Vertices of a hyperbola are at (-5, -2) and (-5, 12) and the point (-5, 30) is one of its foci. What is the equation of its hyperbola

Answer 1

With the provided vertices and the focus point, the hyperbola will look like the one in the picture below.

notice the "c" distance from the center to the focus point.

since it's a vertical hyperbola, the positive fraction will be the one with the "y" in it, its center is clearly half-way between the vertices at -5, 5.

its major axis or traverse axis goes from -2 up to 12, so is 14 units long, therefore the "a" component is half that, or 7.

$\bf \textit{hyperbolas, vertical traverse axis } \\\\ \cfrac{(y- k)^2}{ a^2}-\cfrac{(x- h)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h, k\pm a)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases}\\\\ -------------------------------$

$\bf \begin{cases} h=-5\\ k=5\\ a=7\\ c=25 \end{cases}\implies \cfrac{(y-5)^2}{7^2}-\cfrac{[x-(-5)]^2}{b}=1 \\\\\\ \cfrac{(y-5)^2}{7^2}-\cfrac{(x+5)^2}{b}=1\\\\ -------------------------------\\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b\implies \sqrt{25^2-7^2}=b \\\\\\ \sqrt{625-49}=b\implies \sqrt{576}=b\implies 24=b\\\\ -------------------------------\\\\ \cfrac{(y-5)^2}{7^2}-\cfrac{(x+5)^2}{24^2}=1\implies \cfrac{(y-5)^2}{49}-\cfrac{(x+5)^2}{576}=1$