Question

# (a) Show that the asymptotes of the hyperbola$x ^ { 2 } - y ^ { 2 } = 5$are perpendicular to each other. (b) Find an equation for the hyperbola with foci$( \pm c ,0 )$and with asymptotes perpendicular to each other.

Solution

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(a) $x^2-y^2=5\;\to\;\dfrac{x^2}{5}-\dfrac{y^2}{5}=1$ is hyperbola of the form $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ where $a^2=b^2=5\;\to\;a=b=\sqrt5$.

Its asymptotes are then of the form $y=\pm\dfrac{b}{a}x\;\to\;y=\pm\dfrac{\sqrt5}{\sqrt5}x\;\to\;y=\pm x$.

Recall that perpendicular lines have slopes with a product of $-1$. Since $y=x$ has a slope of 1, $y=-x$ has a slope of $-1$, and the product of the slopes is $1(-1)=-1$, then $x^2-y^2=5$ has perpendicular asymptotes.

(b) If the foci are $(\pm c,0)$, then the equation must be of the form $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ where $c^2=a^2+b^2$ and the asymptotes are $y=\pm\dfrac{b}{a}x$.

The slopes of the asymptotes are $\dfrac{a}{b}$ and $-\dfrac{a}{b}$ so if they are perpendicular, then $-\dfrac{a}{b}\left(\dfrac{a}{b}\right)=-1\;\to\;\dfrac{a^2}{b^2}=1\;\to\;a^2=b^2$.

We then get:

$c^2=a^2+b^2\;\to\;c^2=a^2+a^2\;\to\;c^2=2a^2\;\to\;a^2=\dfrac{c^2}{2}$

Therefore, $b^2=a^2=\dfrac{c^2}{2}$ so the equation is:

$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1\;\to\;\dfrac{x^2}{c^2/2}-\dfrac{y^2}{c^2/2}=1\;\to\;\boxed{x^2-y^2=\dfrac{c^2}{2}}$

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