Answer
Equation: $\frac{x^2}{9}-\frac{y^2}{9}=1$
Foci: $(\pm3\sqrt{2},0)$
Asymptotes: $y=\pm x$
Work Step by Step
Recall: The hyperbola whose the vertices are $(\pm a,0)$ has the equation $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.
In the figure, the vertices are $(\pm 3,0)$.
We get $a=3$ and so $a^2=9$.
The hyperbola also passes through the point $(5,4)$.
Then,
$\frac{5^2}{a^2}-\frac{4^2}{b^2}=1$
$\frac{25}{9}-\frac{16}{b^2}=1$
$\frac{16}{b^2}=\frac{25}{9}-1$
$\frac{16}{b^2}=\frac{16}{9}$
$b^2=9$
So, the equation of the hyperbola is $\frac{x^2}{9}-\frac{y^2}{9}=1$.
Find the value of $c$:
$c=\sqrt{a^2+b^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$
Find the foci:
$(\pm c,0)=(\pm 3\sqrt{2},0)$
Find the asymptotes:
$y=\pm (b/a)x$
$y=\pm (3/3)x$
$y=\pm x$
