Answer
$x^2-\frac{3y^2}{169}=1$
Work Step by Step
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 $(2,13)$:
$\frac{2^2}{1}-\frac{13^2}{b^2}=1$
$-\frac{13^2}{b^2}=1-4=-3$
$-169=-3b^2$
$b^2=\frac{169}{3}$
$\frac{x^2}{1}-\frac{y^2}{\frac{169}{3}}=1$
$x^2-\frac{3y^2}{169}=1$
