Question

Find the coordinates of the points of contact of common tangents to the two hyperbolas $x^2-y^2=3a^2$ and $xy=2a^2$.

Answer 1

Given hyperbolas are

$x^2-y^2=3a^2.......\left(i\right)xy={2a}^2........\left(ii\right)$

Equation of tangent to $\left(i\right)$ at $(\sqrt{3}a\sec \theta ,\sqrt{3}a\tan \theta )$ is

$x\sqrt{3}a\sec \theta -y\sqrt{3}a\tan \theta =3a^2\frac{x}{\sqrt{3}a}-\frac{y}{\sqrt{3}a}\sin \theta =\cos \theta .....\left(iii\right)$

It also touches $\left(ii\right)$

Substituting $x$ from $\left(iii\right)$ in $\left(ii\right)$

$(\sqrt{3}a\cos \theta +y\sin \theta )y={2a}^2{y}^2\sin \theta +\sqrt{3}ay\cos \theta -{2a}^2=\mathrm{0.......}\left(iv\right)$

As the line touches the curve so the roots must be equal

$\Rightarrow D=0{\left(\sqrt{3}a\cos \theta \right)}^2-4(-{2a}^2)\sin \theta =03a^2{\cos }^2\theta +8a^2\sin \theta =03a^2(1-{\sin }^2\theta )+8a^2\sin \theta =03a^2{\sin }^2\theta -8a^2\sin \theta -3a^2=03a^2{\sin }^2\theta -9a^2\sin \theta +a^2\sin \theta -3a^2=03\sin \theta (\sin \theta -3)+(\sin \theta -3)=0(3\sin \theta +1)(\sin \theta -3)=0\Rightarrow \sin \theta =-\frac{1}{3}{\cos }^2\theta =1-{\sin }^2\theta \Rightarrow \cos \theta =\pm \frac{2\sqrt{2}}{3}\Rightarrow \sec \theta =\pm \frac{3}{2\sqrt{2}}{\tan }^2\theta ={\sec }^2\theta -1\Rightarrow \tan \theta =\pm \frac{1}{2\sqrt{2}}$

Substituting $\sin \theta$ and $\cos \theta$ in $\left(iv\right)$

$y^2\sin \theta +\sqrt{3}a\cos \theta y-{2a}^2=0y^2(-\frac{1}{3})+\sqrt{3}ay(\pm \frac{2\sqrt{2}}{3})-{2a}^2=0y^2\pm 2\sqrt{6}ay-6a^2=0\Rightarrow y=\pm \sqrt{6}a$

Substituting $y$ , $\sin \theta$ and $\cos \theta$ in $\left(iii\right)$

$\Rightarrow y=\pm \sqrt{6}a\Rightarrow x=\frac{\pm \sqrt{6}a}{3}$

So the point of contact to hyperbola $\left(ii\right)$ is $(\frac{\pm \sqrt{6}a}{3},\pm \sqrt{6}a)$

Point of contact to $\left(i\right)$ is $(\sqrt{3}a\sec \theta ,\sqrt{3}a\tan \theta )$

Substituting $\tan \theta$ and $\sec \theta$

$\Rightarrow$ point of contact $(\frac{\pm 3\sqrt{6}a}{4},\frac{\pm \sqrt{6}a}{4})$