How do you find the locus of the center of the hyperbola having asymptotes given by y-x tan alpha+1=0 and y-x tan(alpha+pi/4)+2=0, where alpha varies?

Dec 15, 2016

The equations of the locus are:

$x = \cos \alpha \left(\cos \alpha - \sin \alpha\right)$

$y = - 1 + \sin \alpha \left(\cos \alpha - \sin \alpha\right)$

Explanation:

By definition the center of the hyperbola is the point where the asymptotes cross, so that they solve the system:

$y - x \tan \alpha + 1 = 0$
$y - x \tan \left(\alpha + \frac{\pi}{4}\right) + 2 = 0$

$y - x \tan \alpha = - 1$
$y - x \tan \left(\alpha + \frac{\pi}{4}\right) = - 2$

Subtracting member by member:

$x \left(\tan \left(\alpha + \frac{\pi}{4}\right) - \tan \alpha\right) = 1$

So:

$x = \frac{1}{\tan \left(\alpha + \frac{\pi}{4}\right) - \tan \alpha}$

Now note that:

$\frac{1}{\tan \left(\alpha + \frac{\pi}{4}\right) - \tan \alpha} = \frac{1}{\sin \frac{\alpha + \frac{\pi}{4}}{\cos} \left(\alpha + \frac{\pi}{4}\right) - \sin \frac{\alpha}{\cos} \alpha} = \frac{\cos \left(\alpha + \frac{\pi}{4}\right) \cos \alpha}{\sin \left(\alpha + \frac{\pi}{4}\right) \cos \alpha - \sin \alpha \cos \left(\alpha + \frac{\pi}{4}\right)} = \cos \alpha \frac{\cos \alpha \cos \left(\frac{\pi}{4}\right) - \sin \alpha \sin \left(\frac{\pi}{4}\right)}{\sin} \left(\alpha + \frac{\pi}{4} - \alpha\right) = \cos \alpha \left(\cos \alpha - \sin \alpha\right)$

so:

$x = \cos \alpha \left(\cos \alpha - \sin \alpha\right)$

$y = - 1 + \tan \alpha \cos \alpha \left(\cos \alpha - \sin \alpha\right) = - 1 + \sin \alpha \left(\cos \alpha - \sin \alpha\right)$

Are the equations of the locus.

Dec 15, 2016

This locus is the circle, with center at $\left(\frac{1}{2} , - \frac{3}{2}\right)$ and
radius $\frac{1}{\sqrt{2}}$.

Explanation:

The asymptotes meet at the center of the hyperbola.

Let $t = \tan \alpha$. Use $\tan \left(A + B\right) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ and $\tan \left(\frac{\pi}{4}\right) = 1$.

The coordinates (x, y) of the center are given by

$y = x t - 1 = x \frac{t + 1}{1 - t .} - 2$. Eliminating t,

$y = x \left(\frac{1 + \frac{y + 1}{x}}{\left(1 - \frac{y + 1}{x}\right)}\right) - 2$

$= x \left(\frac{x + y + 1}{x - y - 1}\right) - 2$. Cross multiplying,

$x y - {y}^{2} - y = {x}^{2} + x y + x - 2 \left(x - y - 1\right)$. Simplifying,

${x}^{2} + {y}^{2} - x + 3 y + 2 = 0$. In the standard form,

${\left(x - \frac{1}{2}\right)}^{2} + {\left(y + \frac{3}{2}\right)}^{2} = {\left(\frac{1}{\sqrt{2}}\right)}^{2}$

This locus represents the circle with center at $\left(\frac{1}{2} , - \frac{3}{2}\right)$ and

radius $\frac{1}{\sqrt{2}}$.