Question

Write an equation for each hyperbola.

vertices (0,2),(0,-2)
foci (0, +/- square root of 5)

Answer 1

Check the picture below, so the hyperbola looks more or less like so, hmmm so is a vertical one, so is opening towards the y-axis, meaning the fraction that is positive is the one with the y-variable, its traverse axis length is 4 units, from +2 to -2, meaning that a = 2, and the distance either focus to the center is √5, so

`\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 √( a ^2 + b ^2) \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} h=0\\ k=0\\ a=2\\ c=√(5) \end{cases}\qquad \cfrac{(y- 0)^2}{ 2^2}-\cfrac{(x- 0)^2}{ b^2}=1`

`√(5)=√(2^2+b^2)\implies 5=4+b^2\implies 1=b^2\implies √(1)=b\implies 1=b \\\\[-0.35em] ~\dotfill\\\\ \cfrac{(y- 0)^2}{ 2^2}-\cfrac{(x- 0)^2}{ 1^2}=1\implies \cfrac{y^2}{4}~~ - ~~x^2~~ = ~~1`