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

Answer: (C)

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