Question

The line $y=mx+c$ is a normal to $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ then $\frac{a^2}{m^2}-b^2=$

• A. $\frac{(a^2-b^2{)}^2}{2c^2}$
• B. $\frac{(a^2+b^2{)}^2}{4c^2}$
• C. $\frac{(a^2+b^2{)}^2}{c^2}$
• D. $\frac{(a^2-b^2{)}^2}{4c^2}$

Answer 1

Answer: (C)

$\frac{(a^2+b^2{)}^2}{c^2}$

Equation of normal to any point $\theta$ to the hyperbola is given by,
$\frac{ax}{\sec \theta }+\frac{by}{\tan \theta }=a^2+b^2$
$\Rightarrow y=-\frac{a}{b}\sin \theta x+\frac{a^2+b^2}{b}\tan \theta$
Comparing with line $y=mx+c$ we get,
$m=-\frac{a}{b}\sin \theta ,c=\frac{a^2+b^2}{b}\tan \theta$
$\therefore \frac{a}{m^2}-b^2=b^2{\text{cosec}}^2\theta -b^2=b^2{\cot }^2\theta =\frac{(a^2+b^2{)}^2}{c^2}$