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

$\begin{array}{1 1}(a)\;\large\frac{xx_1}{a^2}+\frac{yy_1}{b^2}\normalsize=1\\(b)\;\large\frac{xx_1}{a^2}-\frac{yy_1}{b^2}\normalsize=1\\(c)\;\large\frac{xx_1}{a^3}+\frac{yy_1}{b^3}\normalsize=1\\(d)\;\text{None of these}\end{array}$

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1 Answer

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Let us suppose a point P(x,y) at which we find the equation of tangent
Let us find the slope at $(x_1,y_1)$
Slope =$\large\frac{2x}{a^2}+\frac{2yy'}{b^2}$$=0$
$y'=-\large\frac{-x_2\times b^2}{a^2\times y_1}$
Hence equation of tangent at $(x_1,y_1)$ is
$(y-y_1)=-\large\frac{-x_1\times b^2}{a^2\times y'}$$(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}$$=1$
Hence (a) is the correct answer.
answered Feb 7, 2014 by sreemathi.v
 

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