Answer
(a) See explanations.
(b) $x^2-y^2=\frac{c^2}{2}$
Work Step by Step
(a) Two lines will be perpendicular if one slope is the negative reciprocal of the other. For the hyperbola $x^2-y^2=5$, divide 5 on both sides of the equation we have $\frac{x^2}{(\sqrt 5)^2} -\frac{y^2}{(\sqrt 5)^2}=1 $. Thus the asymptotes are $y=\pm\frac{(\sqrt 5)}{(\sqrt 5)}x=\pm x$. Use the perpendicular test rule stated earlier, we can see that these two asymptotes are perpendicular to each other.
(b) Assume a standard hyperbola equation $\frac{x^2}{a^2} -\frac{y^2}{b^2}=1 $, we have the asymptotes as $y=\pm\frac{b}{a}x$. Knowing that these two asymptotes are perpendicular to each other, we have $\frac{b}{a}=\frac{a}{b}$ which gives $a^2=b^2$. Use the formula $c^2=a^2+b^2$, we have $2a^2=c^2$, so that $a^2=b^2=\frac{c^2}{2}$. Thus the equation will be $\frac{2x^2}{c^2} -\frac{2y^2}{c^2}=1 $ or $x^2-y^2=\frac{c^2}{2}$
