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

Answer 1

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

$\frac{x^2}{24}-\frac{y^2}{18}=1\Rightarrow a=\sqrt{24}$ and $b=\sqrt{18}$
We know parametric form of normal to hyperbola is
$ax\cos \theta +by\cot \theta =a^2+b^2\Rightarrow \sqrt{24}x\cos \theta +\sqrt{18}y\cot \theta =42Atx=0,y=\frac{42}{\sqrt{18}\cdot \cot \theta }=\frac{42}{\sqrt{18}}\tan \theta \cdots \left(i\right)Fory=mx+7\sqrt{3}y=7\sqrt{3}\cdots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$
$\frac{42}{\sqrt{18}}\tan \theta =7\sqrt{3}\Rightarrow \tan \theta =\sqrt{\frac{3}{2}}\Rightarrow \sin \theta =\pm \sqrt{\frac{3}{5}}$
Also, slope of parametrical normal
$-\frac{a\cos \theta }{b\cot \theta }=m\Rightarrow m=-\sqrt{\frac{4}{3}}\sin \theta =-\frac{2}{\sqrt{5}}$
or $\frac{2}{\sqrt{5}}$