Question:
If the line $y=m x+7 \sqrt{3}$ is normal to the hyperbola $\frac{x^{2}}{24}-\frac{y^{2}}{18}=1$, then a value of $\mathrm{m}$ is :
Correct Option: , 3
Solution:
Since, $1 \mathrm{x}+\mathrm{my}+\mathrm{n}=0$ is a normal to $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$,
then $\frac{a^{2}}{l^{2}}-\frac{b^{2}}{m^{2}}=\frac{\left(a^{2}+b^{2}\right)^{2}}{n^{2}}$
but it is given that $m x-y+7 \sqrt{3}$ is normal to hyperbola
$\frac{x^{2}}{24}-\frac{y^{2}}{18}=1$
then $\frac{24}{m^{2}}-\frac{18}{(-1)^{2}}=\frac{(24+18)^{2}}{(7 \sqrt{3})^{2}} \Rightarrow m=\frac{2}{\sqrt{5}}$