Question

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

• A. $\pm \frac{2}{\sqrt{3}}$
• B. $\sqrt{3}$
• C. $-\frac{\sqrt{3}}{2}$
• D. none of these

Answer 1

The correct option is A $\pm \frac{2}{\sqrt{3}}$

Given hyperbola is, $\frac{x^2}{16}-\frac{y^2}{9}=1$
$a^2=16,b^2=9$
and the line is $y=mx+\frac{25}{\sqrt{3}}$
$c=\frac{25}{\sqrt{3}}$ and slope $m=m$
Now using condition of normality $c^2=\frac{(a^2+b^2{)}^2{m}^2}{a^2-b^2{m}^2}$
$\Rightarrow \frac{{25}^2}{3}=\frac{{25}^2{m}^2}{16-9m^2}$
$\Rightarrow m=\pm \frac{2}{\sqrt{3}}$