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

Find the equation of normal for ellipse $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$?

$\begin{array}{1 1}(a)\;\large\frac{a^2x}{x_1}+\frac{b^2y}{y_1}\normalsize= a^2-b^2\\(b)\;\large\frac{a^2x}{x_1}-\frac{b^2y}{y_1}\normalsize= a^2-b^2\\(c)\;\large\frac{a^2x}{x_1}+\frac{b^3y}{y_1}\normalsize =a^3-b^3\\(d)\;\text{None of these}\end{array}$

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1 Answer

We know slope of tangent at $(x_1,y_1)$ is $-\large\frac{x_1b^2}{a^2y_1}$
Slope of normal $\large\frac{a^2y_1}{x_1b^2}$
Hence equation of normal is
$(y-y_1)=\large\frac{a^2y_1}{x_1b^2}$$(x-x_1)$
$\large\frac{y-y_1}{a^2y_1}=\frac{x}{x_1b^2}-\frac{1}{b^2}$
$\large\frac{y}{a^2y_1}-\frac{1}{a^2}=\frac{x}{x_1b^2}-\frac{1}{b^2}$
$\large\frac{a^2x}{x_1}-\frac{b^2y}{y_1}=$$a^2-b^2$
Hence (b) is the correct answer.
answered Feb 7, 2014 by sreemathi.v
 

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