Browse Questions

Find the equation of the tangent and normal to the hyperbola $\large\frac{x^{2}}{9}-\frac{y^{2}}{12}=$$1 at \theta=\large\frac{\pi}{6} Can you answer this question? 1 Answer 0 votes Toolbox: • The equation of tangent at \theta is given by \large\frac{x\sec \theta}{a}-\large\frac{y\tan \theta}{b}$$=1$
• The equation of normal at $\theta$ is given by $\large\frac{ax}{\sec \theta}+\large\frac{by}{\tan \theta}=$$a^2+b^2 Step 1: \large\frac{x^2}{9}-\frac{y^2}{12}$$=1$
$\Rightarrow a^2=9$ and $b^2=12$
$\Rightarrow a=3,b=2\sqrt 3$
The parametric equations are $x=a\sec\theta,y=b\tan\theta$
(i.e) $x=3\sec\theta,y=2\sqrt 3\tan\theta$
Step 2:
The tangent at $\theta$ is $\large\frac{x\sec \theta}{a}-\large\frac{y\tan \theta}{b}$$=1 a=3,b=2\sqrt 3,\theta=\large\frac{\pi}{6} \large\frac{x\sec \Large\frac{\pi}{6}}{3}-\large\frac{y\tan \Large\frac{\pi}{6}}{2\sqrt 3}$$=1$
We know that $\sec\large\frac{\pi}{6}=\large\frac{2}{\sqrt 3}$ and $\tan\large\frac{\pi}{6}=\large\frac{1}{\sqrt 3}$
$\large\frac{x.2}{3\sqrt 3}-\large\frac{y.\Large\frac{1}{\sqrt 3}}{2\sqrt 3}$$=1 \large\frac{2x}{3\sqrt 3}-\large\frac{y}{6}$$=1$
$12x-3\sqrt 3y=18\sqrt 3$
The above equation is divided by $3$ we get,
$4x-\sqrt 3y=6\sqrt 3$
Step 3:
The normal at $\theta$ is $\large\frac{ax}{\sec \theta}+\large\frac{by}{\tan \theta}=$$a^2+b^2 \large\frac{3x}{\sec \Large\frac{\pi}{6}}+\large\frac{2\sqrt 3y}{\tan \Large\frac{\pi}{6}}=$$3^2+(2\sqrt 3)^2$
We know that $\sec\large\frac{\pi}{6}=\large\frac{2}{\sqrt 3}$ and $\tan\large\frac{\pi}{6}=\large\frac{1}{\sqrt 3}$
$\large\frac{3\sqrt 3x}{2}$$+6y=21$
$3\sqrt 3x+12y=42$
The above equation is divided by $3$ we get,
$\sqrt 3x+4y=14$
Step 4:
Equation of tangent : $4x-\sqrt 3y=6\sqrt 3$
Equation of normal : $\sqrt 3x+4y=14$