logo

Ask Questions, Get Answers

X
 
Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Coordinate Geometry

Find the equation of normal for the hyperbola $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1$ in parametric form?

$\begin{array}{1 1}(a)\;\large\frac{by}{\tan \theta}+\frac{ax}{\sec\theta}\normalsize =a^2+b^2\\(b)\;\large\frac{by}{\tan \theta}-\frac{ax}{\sec\theta}\normalsize =a^2+b^2\\(c)\;\large\frac{by}{\tan \theta}+\frac{ax}{\sec\theta}\normalsize =a^2-b^2\\(d)\;\text{None of these}\end{array}$

1 Answer

Equation of normal at $(x_1,y_1)$ is
$\large\frac{b^2y}{y_1}+\frac{a^2x}{x_1}$$=a^2+b^2$
Now replacing $(x_1,y_1)$ by $(a\sec\theta,b\tan\theta)$
Hence $\large\frac{b^2y}{b\tan\theta}+\frac{a^2x}{a\sec\theta}$$=a^2+b^2$
$\large\frac{by}{\tan\theta}+\frac{ax}{\sec\theta}$$=a^2+b^2$
Hence (a) is the correct answer.
answered Feb 10, 2014 by sreemathi.v
 

Related questions

Download clay6 mobile appDownload clay6 mobile app
...
X