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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Find the equation of tangent for the hyperbola $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1$?

$\begin{array}{1 1}(a)\;\large\frac{xx_1}{a^2}-\frac{yy_1}{b^2}\normalsize =1\\(b)\;\large\frac{xx_1}{a^2}+\frac{yy_1}{b^2}\normalsize =1\\(c)\;\large\frac{xx_1}{b^2}-\frac{yy_1}{a^2}\normalsize =1\\(d)\;\text{None of these}\end{array}$

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1 Answer

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Let us consider a point $(x_1,y_1)$ at which we should find the tangent
Hence slope at point $(x_1,y_1)$ is
$\large\frac{2x}{a^2}-\frac{2yy_1}{b^2}$$=0$
$\large\frac{2x}{a^2}=\frac{yy_1}{b^2}$
At $(x_1,y_1)$
$\large\frac{x_1b^2}{y_1a^2}$$=y'$
Hence equation of tangent is
$(y-y_1)=\large\frac{x_1b^2}{y_1a^2}$$(x-x_1)$
$\large\frac{yy_1}{b^2}-\frac{y_1^2}{b^2}=\frac{xx_1}{a^2}-\frac{x_1^2}{a^2}$
$\large\frac{xx_1}{a^2}-\frac{yy_1}{b^2}\normalsize =1$
Hence (a) is the correct answer.
answered Feb 10, 2014 by sreemathi.v
 

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