Question

Tangents are drawn to hyperbola $\frac{x^2}{25}-\frac{y^2}{9}=1$ at points whose eccentric angles are $30°$ and $60°$. Find point of intersection of tangents at these points.

Answer 1

Given: Hyperbola $\frac{x^2}{25}-\frac{y^2}{9}=1$; Eccentric angles: $\alpha =30°$ and $\beta =60°$

To Find: Point of intersection of tangents

Step - $1$: Recall equation of chord of contact and equation of chord joining points $\alpha$ and $\beta$.

Step - $2$: Find the equation of both the chords.

Step - $3$: Equate both chords to get point of intersection.

Let the point of intersection be $(h,k)$.

$\frac{x^2}{25}-\frac{y^2}{9}=1\Rightarrow a=5;b=3$

We know that, the equation of chord of contact to hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ from point $P(h,k)$ is $\frac{hx}{a^2}-\frac{ky}{b^2}=1$.

$\Rightarrow$ Chord of contact $PQ$ is: $\frac{hx}{25}-\frac{ky}{9}=1\cdots \left(1\right)$

Equation of chord joining points $P\left(\alpha \right)$ and $Q\left(\beta \right)$ is:
$\frac{x}{a}\cos (\frac{\alpha -\beta }{2})-\frac{y}{b}\sin (\frac{\alpha +\beta }{2})=\cos (\frac{\alpha +\beta }{2})$

$\Rightarrow$ Chord of contact $PQ$ is:
$\frac{x}{5}\cos (\frac{30°-60°}{2})-\frac{y}{3}\sin (\frac{30°+60°}{2})=\cos (\frac{30°+60°}{2})$

$\Rightarrow \frac{x}{5}\cos (-15°)-\frac{y}{3}\sin \left(45°\right)=\cos \left(45°\right)$

$\Rightarrow \frac{x}{5}\cos \left(15°\right)-\frac{y}{3\sqrt{2}}=\frac{1}{\sqrt{2}}$

$\Rightarrow \frac{x}{5}(\frac{\sqrt{3}+1}{2\sqrt{2}})-\frac{y}{3\sqrt{2}}=\frac{1}{\sqrt{2}}$

$\Rightarrow \frac{x(\sqrt{3}+1)}{10}-\frac{y}{3}=1\cdots \left(2\right)$

Equation $\left(1\right)$ and $\left(2\right)$ represent the same line.
$\therefore \frac{\frac{h}{25}}{\frac{\sqrt{3}+1}{10}}=\frac{-\frac{k}{9}}{-\frac{1}{3}}=1$

$\Rightarrow \frac{2h}{5(\sqrt{3}+1)}=\frac{k}{3}=1$

$\therefore \frac{2h}{5(\sqrt{3}+1)}=1\text{and}\frac{k}{3}=1$

$\Rightarrow h=\frac{5(\sqrt{3}+1)}{2}\text{and}k=3$

Point of intersection is: $(\frac{5(\sqrt{3}+1)}{2},3)$.