Question

Find the equation of tangent to curve $x^2-9y^2=9$ at the point whose eccentric angle is $\frac{\pi }{3}$ .

• A. $2x-3\sqrt{3}y=3$
• B. $x-\sqrt{3}y=6$
• C. $x-3\sqrt{3}y=3$
• D. $2x-3\sqrt{3}y=6$

Answer 1

The correct option is A $2x-3\sqrt{3}y=3$

The equation of hyperbola is $x^2-9y^2=9$.
$\Rightarrow \frac{x^2}{9}-\frac{9y^2}{9}=1$
$\Rightarrow \frac{x^2}{9}-y^2=1$
Comparing with standard form of hyperbola i.e.,
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
we get $a=3,b=1$

The equation of tangent at point $P\left(\theta \right)$: $\frac{x}{a}\sec \theta -\frac{y}{b}\tan \theta =1$
Substituting values of $a,b$ and $\theta$ in the equation of tangent:
$\frac{x}{3}\sec \frac{\pi }{3}-\frac{y}{1}\tan \frac{\pi }{3}=1$
$\Rightarrow \frac{x}{3}\left(2\right)-\frac{y}{1}\left(\sqrt{3}\right)=1$
$\Rightarrow 2x-3\sqrt{3}y=3$
Hence, $2x-3\sqrt{3}y=3$ is the required equation.