Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
0 votes

Find the equation of normal in slope form for the hyperbola?

$\begin{array}{1 1}(a)\;y=mx\pm\large\frac{m(a^2+b^2)}{\sqrt{a^2+m^2b^2}}\\(b)\;y=mx\mp\large\frac{m(a^2+b^2)}{\sqrt{a^2-m^2b^2}}\\(c)\;y=mx\pm\large\frac{m(a^2-b^2)}{\sqrt{a^2+m^2b^2}}\\(d)\;\text{None of these}\end{array}$

Can you answer this question?

1 Answer

0 votes
The equation of normal to the hyperbola $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1$ at $(x_1,y_1)$ is
'm' is the slope of the normal then
Since $(x_1,y_1)$ lies on $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1$
Hence $\large\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}$$=1$
$x_1=\pm \large\frac{a^2}{\sqrt{(a^2-b^2m^2)}}$
Equation of normal
$y-\big(\mp\large\frac{mb^2}{\sqrt{a^2-m^2b^2}}\big)=$$m\big(x-\big(\pm \large\frac{a^2}{\sqrt{a^2-m^2b^2}}\big)\big)$
Hence (b) is the correct answer.
answered Mar 26, 2014 by balaji.thirumalai

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App