If the line y=mx+73 is normal to the hyperbola x224−y218=1 then a value of m is :

A
$2 / \sqrt{5}$
B
$3 / \sqrt{5}$
C
$\sqrt{15} / 2$
D
$\sqrt{5} / 2$

Solution:

Given hyperbola is $\frac{x^{2}}{24} - \frac{y^{2}}{18} = 1$

$\Rightarrow a = \sqrt{24}, b = \sqrt{18}$

Now equation of normal at $(a \sec θ, b \tan θ)$ is

$a x \cos θ + b y \cot θ = a^{2} + b^{2}$

$\begin{array}{l} \Rightarrow \sqrt{24} x \cos 0 + \sqrt{18} y \cot θ = 24 + 18 \\ \Rightarrow \sqrt{24} \cos θ x + \sqrt{18} \cot θ y = 42 \end{array}$

At $x = 0, y = \frac{42}{\sqrt{18} \cot θ} = 7 \sqrt{3}$

[ $∵$ Given that normal is $y = m x + 7 \sqrt{3}$ ]

$\Rightarrow \tan θ = \sqrt{\frac{3}{2}} \Rightarrow \sin θ = \pm \sqrt{\frac{3}{5}}$ ... (i)

Slope of normal $= \frac{- \sqrt{24} \cos θ}{\sqrt{18} \cot θ} = m$

$\Rightarrow m = - \sqrt{\frac{4}{3}} \sin θ = - \frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{5}}$ or $\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{5}}$ [Using (i)]

$= - \frac{2}{\sqrt{5}}$ or $\frac{2}{\sqrt{5}}$

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 m is :

Solution:

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