JEE Main 2019 — Hyperbola Question with Solution

The given equation of the hyperbola is $\frac{x^2}{24}-\frac{y^2}{18}=1,$ which is of the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$

$\Rightarrow a^2=24; b^2=18$

We know that the equation of a normal of slope $m$ to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $y=mx\pm \frac{m\left(a^2+b^2\right)}{\sqrt{a^2-b^2{m}^2}}$

Thus, the equation of the normal to the given hyperbola is $y=mx\pm \frac{m\left(24+18\right)}{\sqrt{24-18m^2}}$

$y=mx\pm \frac{m\left(42\right)}{\sqrt{24-18m^2}}$

Given, the normal is $y=mx+7\sqrt{3}$

Hence, comparing the two equations, we get $\frac{42m}{\sqrt{24-18m^2}}=7\sqrt{3}$

$\Rightarrow \frac{{\left(42m\right)}^2}{24-18m^2}=49\times 3$

$\Rightarrow 42\times 42m^2=49\times 3\left(24-18m^2\right)$

$\Rightarrow 12m^2=24-18m^2$

$\Rightarrow m^2=\frac{24}{30}=\frac{4}{5}$

$\Rightarrow m=\pm \sqrt{\frac{4}{5}}=\pm \frac{2}{\sqrt{5}}.$