Answer
$\frac{x^2}{16}-\frac{y^2}{16}=1$
Work Step by Step
Asymptotes:
$y=±\frac{a}{b}x$ (vertical transverse axis)
$y=±\frac{b}{a}x$ (horizontal transverse axis)
$y=±x$
$a=b$
The point $(5,3)$ is above $y=-x$ and below $y=x$ because $5\gt3$ (see the graph). We have a hyperbola with horizontal transverse axis:
Use the point $(5,3)$.
Hyperbola with horizontal transverse axis:
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\frac{5^2}{a^2}-\frac{3^2}{a^2}=1$
$\frac{16}{a^2}=1$
$a^2=16$
$a=4$
Finally:
$\frac{x^2}{4^2}-\frac{y^2}{4^2}=1$
$\frac{x^2}{16}-\frac{y^2}{16}=1$
