Find the locus of points where a pair tangents drawn on the hyperbola $x^2 – y^2 = a^2 $ enclose an angle of $45$ degrees. This is what I've done so far.
$\theta$ between tangents is $45$ so
$\lvert \frac{m_2-m_1}{1+m_1m_2)}\rvert = tan45 =1$
$\lvert m_2-m_1\rvert = \lvert 1+m_1m_2\rvert$
I think point-slope form of hyperbola tangent will also be helpful here but I'm not sure how to apply it:
$$y = mx \pm \sqrt{a^2m^2-b^2}$$
Best Answer
Hint :
The equation for pair of tangents can be converted into a quadratic in $m$ :
$$(x^2-a^2)m^2-2xy \cdot m + y^2+b^2=0$$
Now use Vieta's formulas on the obtained condition for $m$, $$(m_1+m_2)^2-4m_1m_2=(m_1-m_2)^2=(1+m_1m_2)^2$$
to obtain the desired locus.