Algebra and Trigonometry 10th Edition

$x^2-\frac{3y^2}{169}=1$
Vertices: $(±a,0)=(±1,0)$ $a=1$ Hyperbola with horizontalal transverse axis: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ $\frac{x^2}{1}-\frac{y^2}{b^2}=1$ Passes through the point $(2,13)$: $\frac{2^2}{1}-\frac{13^2}{b^2}=1$ $-\frac{13^2}{b^2}=1-4=-3$ $-169=-3b^2$ $b^2=\frac{169}{3}$ $\frac{x^2}{1}-\frac{y^2}{\frac{169}{3}}=1$ $x^2-\frac{3y^2}{169}=1$