Question

Find the equation of normal to the point on the hyperbola $S\equiv 6x^2-4y^2=3$ whose eccentric angle is $\frac{\pi }{6}$.

• A. $\sqrt{6}x+2y=5$
• B. $\sqrt{6}x-6y=5$
• C. $\sqrt{6}x+6y=5$
• D. $6x+6y=5$

Answer 1

The correct option is C $\sqrt{6}x+6y=5$

Given: Hyperbola $S\equiv 6x^2-4y^2=3$; Eccentric angle is $\frac{\pi }{6}$

To Find: Equation of normal

Step - $1$: Recall the equation of normal in parametric form.

Step - $2$: Find $a$ and $b$.

Step - $3$: Substitute the values and simplify.

We have the hyperbola, $6x^2-4y^2=3$

$\Rightarrow \frac{6x^2}{3}-\frac{4y^2}{3}=1$

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

Here, $a=\frac{1}{\sqrt{2}},b=\frac{\sqrt{3}}{2}$

Equation of normal in parametric form is $\frac{ax}{\sec \theta }+\frac{by}{\tan \theta }=a^2+b^2$

$\therefore$ Equation of normal:
$\frac{(\frac{1}{\sqrt{2}})x}{\sec (\frac{\pi }{6})}+\frac{(\frac{\sqrt{3}}{2})y}{\tan (\frac{\pi }{6})}=\frac{1}{2}+\frac{3}{4}$

$\Rightarrow \frac{(\frac{1}{\sqrt{2}})x}{(\frac{2}{\sqrt{3}})}+\frac{(\frac{\sqrt{3}}{2})y}{(\frac{1}{\sqrt{3}})}=\frac{5}{4}$

$\Rightarrow \frac{\sqrt{3}x}{2\sqrt{2}}+\frac{3y}{2}=\frac{5}{4}$

Multiplying the equation by $4$, we get,
$\Rightarrow \sqrt{6}x+6y=5$