Question

The value(s) of $m$, for which the line $y=mx+\frac{25\sqrt{3}}{3}$ , is a normal to the conic $\frac{x^2}{16}-\frac{y^2}{9}=1$ is/are

• A. $\frac{\sqrt{3}}{2}$
• B. $-\frac{2}{\sqrt{3}}$
• C. $-\frac{\sqrt{3}}{2}$
• D. $\frac{2}{\sqrt{3}}$

Answer 1

Answer: (B), (D)

$\frac{2}{\sqrt{3}}$

We know that the equation of the normal of the conic $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at point $(a\sec \theta ,b\tan \theta )$ is
$ax\cos \theta +by\cot \theta =a^2+b^2$ or
$y=\frac{-ax}{b}\sin \theta +\frac{a^2+b^2}{b\cot \theta }$

Comparing above equation with the equation $y=mx+\frac{25\sqrt{3}}{3}$ and taking $a=4,b=3$, we get,
$\frac{a^2+b^2}{b\cot \theta }=\frac{25\sqrt{3}}{3}$
$\Rightarrow \frac{16+9}{3\cot \theta }=\frac{25\sqrt{3}}{3}$
$\Rightarrow \tan \theta =\sqrt{3}$
$\Rightarrow \sin \theta =\frac{\sqrt{3}}{2}$ or $-\frac{\sqrt{3}}{2}$
$m=-\frac{a}{b}\sin \theta =-\frac{2}{\sqrt{3}}\text{or}\frac{2}{\sqrt{3}}$

Alternate solution
Equation of normal to hyperbola in slope form is
$y=mx\pm \frac{(a^2+b^2)m}{\sqrt{a^2-b^2{m}^2}}$
Comparing with the given equation, we get
$\frac{25\sqrt{3}}{3}=\frac{(a^2+b^2)m}{\sqrt{a^2-b^2{m}^2}}$
$\Rightarrow \frac{25m}{\sqrt{16-9m^2}}=\frac{25\sqrt{3}}{3}$
$\Rightarrow 3m^2=16-9m^2$
$\Rightarrow m^2=\frac{16}{12}=\frac{4}{3}$
$\therefore m=\pm \frac{2}{\sqrt{3}}$