Question

If the line $x+y+k=0$ is a normal to the hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$, then $k$ is

• A. $\pm \frac{\sqrt{5}}{13}$
• B. $\pm \frac{13}{\sqrt{5}}$
• C. $\pm \frac{13}{5}$
• D. $\pm \frac{5}{13}$

Answer 1

The correct option is B $\pm \frac{13}{\sqrt{5}}$

The line $lx+my+n=0$ is a normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Then $\frac{x^2}{l^2}-\frac{y^2}{m^2}=\frac{{(a^2+b^2)}^2}{n^2}$
Given hyperbola is $\frac{x^2}{9}-\frac{y^2}{4}=1$
$\Rightarrow \frac{9}{1}-\frac{4}{1}=\frac{{(9+4)}^2}{k^2}$
$\Rightarrow 9-4=\frac{{\left(13\right)}^2}{k^2}$
$\Rightarrow k^2=\frac{169}{5}$

$\Rightarrow k=\pm \frac{13}{\sqrt{5}}$