Question

The equation of normal to the hyperbola $3x^2-y^2=1$ having slope $\frac{1}{3}$ is

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

Answer 1

Answer: (A), (B)

$x-3y=-2\sqrt{2}$

$3x^2-y^2=1$
$\Rightarrow \frac{x^2}{\left(\frac{1}{3}\right)}-y^2=1$
$a^2=\left(\frac{1}{3}\right)$, $b^2=1$
The equation of normal at any point $(\frac{1}{\sqrt{3}}\sec \theta ,\tan \theta )$ on the hyperbola is given by
$\frac{\frac{1}{\sqrt{3}}x}{\sec \theta }+\frac{y}{\tan \theta }=\frac{4}{3}$

$\frac{x}{\sqrt{3}}\cos \theta +y\cot \theta =\frac{4}{3}$

Slope = $-\frac{\cos \theta }{\sqrt{3}}\times \frac{\sin \theta }{\cos \theta }\Rightarrow \frac{-\sin \theta }{\sqrt{3}}=\frac{1}{3}$
$\sin \theta =-\frac{1}{\sqrt{3}}$
$\Rightarrow \theta$ is in $3^{\text{rd}}$ or $4^{\text{th}}$ quadrant.
If $\theta$ is in $3^{\text{rd}}$ quadrant $\Rightarrow \tan \theta =\frac{1}{\sqrt{2}},\sec \theta =\frac{-\sqrt{3}}{\sqrt{2}}$
If $\theta$ is in $4^{\text{th}}$ quadrant $\Rightarrow \tan \theta =\frac{-1}{\sqrt{2}},\sec \theta =\frac{\sqrt{3}}{\sqrt{2}}$
The equations of normals are
$\frac{\sqrt{2}x}{3}-\sqrt{2}y=\frac{4}{3}$
or $\frac{-\sqrt{2}x}{3}+\sqrt{2}y=\frac{4}{3}$
$\Rightarrow x-3y=2\sqrt{2}$
or $x-3y=-2\sqrt{2}$