Chapter 7, Conic Sections - Section 7.3 - Hyperbolas - 7.3 Exercises - Page 585: 28

$\frac{y^2}{144}-\frac{x^2}{25}=1$

Vertices: $(0,±a)=(0,±12)$ $a=12$ Foci: $(0,±c)=(0,±13)$ $c^2=a^2+b^2$ $b^2=c^2-a^2=13^2-12^2=169-144=25$ $b=5$ Asymptotes: $y=±\frac{a}{b}x$ $y=±\frac{12}{5}x$ Hyperbola with vertical transverse axis: $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ $\frac{y^2}{12^2}-\frac{x^2}{5^2}=1$ $\frac{y^2}{144}-\frac{x^2}{25}=1$