If the line $5 x-2 y-6=0$ is a tangent to the hyperbola $5 x^2-k y^2=12$. then the equation of the normal to…
If the line $5 x-2 y-6=0$ is a tangent to the hyperbola $5 x^2-k y^2=12$. then the equation of the normal to this hyperbola at the point $(\sqrt{6}, p)(p \lt 0)$ is
$\sqrt{6} x+2 y=0$
$2 \sqrt{6} x+3 y=3$
$\sqrt{6} x-5 y=21$
$3 \sqrt{6} x-y=21$
Solution
Given equation of hyperbola
$5 x^2-k y^2=12$
$\Rightarrow \frac{x^2}{\frac{12}{5}}-\frac{y^2}{\frac{12}{k}}=1$
So equation of tangent at $\left(x_1, y_1\right)$ is
$\begin{aligned} & \frac{x_1}{a^2} x-\frac{y_1}{b^2} y=1 \\ & \Rightarrow \frac{x_1}{\frac{12}{5}} x-\frac{y_1}{\frac{12}{k}} y=1 \Rightarrow \frac{5 x_1}{12} x-\frac{k y_1}{12} y=1\end{aligned}$
$\Rightarrow \frac{5 x_1}{2} x-\frac{k y_1}{2} y=6$
So, $\frac{5 x_1}{2}=5 \Rightarrow x_1=2, \frac{k y_1}{2}=2 \Rightarrow y_1=\frac{4}{k}$
$\begin{aligned} & \text { Now, } 5 \times 4-k \times \frac{16}{k^2}=12 \\ & \Rightarrow 20-12=\frac{16}{k} \Rightarrow k=2\end{aligned}$
Since, $(\sqrt{6}, p)$ satisfied the hyperbola
$\Rightarrow 5 \times 6-2 \times p^2=12 \Rightarrow p=-3, p \lt 0$
So, equation of normal at $(\sqrt{6},-3)$ is
$\frac{a^2}{x_1} x+\frac{b^2}{y_1} y=a^2+b^2$
$\begin{aligned} & \Rightarrow \frac{12}{5 \times \sqrt{6}} x+\frac{12}{2 \times(-3)} y=\frac{12}{5}+\frac{12}{2} \\ & \Rightarrow \frac{x}{5 \sqrt{6}}-\frac{y}{6}=\frac{1}{5}+\frac{1}{2}\end{aligned}$
$\Rightarrow \frac{\sqrt{6} x-5 y}{30}=\frac{7}{10} \Rightarrow \sqrt{6} x-5 y=21$