Question

If the line $y=mx+7\sqrt{3}$ is normal to the hyperbola $\frac{x^2}{24}-\frac{y^2}{18}=1$, then a value of m is?

• A. $\frac{\sqrt{5}}{2}$
• B. $\frac{3}{\sqrt{5}}$
• C. $\frac{2}{\sqrt{5}}$
• D. $\frac{\sqrt{15}}{2}$

Answer 1

The correct option is C $\frac{2}{\sqrt{5}}$

$\frac{x^2}{24}-\frac{y^2}{18}=1\Rightarrow a=\sqrt{24};b=\sqrt{18}$
Parametric normal:
$\sqrt{24}\cos \theta \cdot x+\sqrt{18}\cdot y\cot \theta =42$
At $x=0:y=\frac{42}{\sqrt{18}}\tan \theta =7\sqrt{3}$ (from given equation)
$\Rightarrow \tan \theta =\sqrt{\frac{3}{2}}$
$\Rightarrow \sin \theta =\pm \sqrt{\frac{3}{5}}$
Slope of parametric normal $=\frac{-\sqrt{24}\cos \theta }{\sqrt{18}\cot \theta }=m$
$\Rightarrow m=-\sqrt{\frac{4}{3}}\sin \theta =-\frac{2}{\sqrt{5}}$ or $\frac{2}{\sqrt{5}}$.