Hyperbola $\displaystyle x^2-y^2+ax+by=0$ and circle $\displaystyle x^2+y^2=a^2+b^2$ intersect at four points. Show that three of them are vertices of an equilateral triangle.
Hello atreyyu
Let $\displaystyle x=r\cos \theta$ and $\displaystyle y=r\sin\theta$, where $\displaystyle r^2=a^2+b^2$.
Then $\displaystyle (x,y)$ lies on the circle for all $\displaystyle \theta$.
Circle intersects hyperbola where:
$\displaystyle r^2\cos^2\theta - r^2\sin^2\theta = -ar\cos\theta-br\sin\theta$
$\displaystyle \implies \cos^2\theta - \sin^2\theta = -(\frac{a}{r}\cos\theta+\frac{b}{r}\sin\theta)$
Now express the LHS as $\displaystyle \cos 2\theta$, and the RHS as $\displaystyle -\cos(\theta -\alpha)$, where $\displaystyle \tan\alpha=...$
Then show that there are three values of $\displaystyle \theta$ that differ by $\displaystyle \frac{2\pi}{3}$ and $\displaystyle \frac{4\pi}{3}$.
Can you do it now?
Grandad