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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Find the equation of normal in slope form for the hyperbola?

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

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1 Answer

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The equation of normal to the hyperbola $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1$ at $(x_1,y_1)$ is
$\large\frac{a62x}{x_1}+\frac{b^2y}{y_1}$$=a^2+b^2$
'm' is the slope of the normal then
$m=-\large\frac{a^2y_1}{b^2x_1}$
$y_1=-\large\frac{b^2x_1m}{a^2}$
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$
$\large\frac{x_1^2}{a^2}-\frac{b^4x_1^2m^2}{a^4b^2}$$=1$
$x_1=\pm \large\frac{a^2}{\sqrt{(a^2-b^2m^2)}}$
$y_1=\mp\large\frac{mb^2}{\sqrt{(a^2-b^2m^2)}}$
Equation of normal
$y-\big(\mp\large\frac{mb^2}{\sqrt{a^2-m^2b^2}}\big)=$$m\big(x-\big(\pm \large\frac{a^2}{\sqrt{a^2-m^2b^2}}\big)\big)$
$y=mx\mp\large\frac{m(a^2+b^2)}{\sqrt{a^2-m^2b^2}}$
Hence (b) is the correct answer.
answered Mar 26, 2014 by balaji.thirumalai
 

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