# Chapter 4 - 4.3 - Conics - 4.3 Exercises - Page 339: 81a

$x^2-\frac{3y^2}{169}=1$

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

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