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

$\begin{array}{1 1}(a)\;\large\frac{a^2}{x_1}\normalsize -x_1,(1-e^2)x_1\\(b)\;\large\frac{a}{x_1}\normalsize- x_1,(1-e^4)x_1\\(c)\;\large\frac{a^2}{x_1}\normalsize+ x_1,(1+e^2)x_1\\(d)\;\text{None of these}\end{array}$

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Equation of tangent at $P(x_1,y_1)$ is
$\large\frac{xx_1}{a^2}+\frac{yy_1}{b^2}$$=1$-----(1)
T lies on x-axis
Put $y=0$ in equ(1)
$x=cT$
CT=$\large\frac{a^2}{x_1}$
$CN=x_1$
Length of sub-tangent $NT=CT-CN$
$\Rightarrow \large\frac{a^2}{x_1}$$-x_1$
Equation of normal at $P(x_1,y_1)$ to the ellipse $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$ is
$\large\frac{x-x_1}{x_1/a^2}=\frac{y-y_1}{y_1/b^2}$-----(2)
$G$ lies on x-axis put $y=0$ in (2)
$\Rightarrow x=CG$
$CG=x_1-\large\frac{b^2}{a^2}$$x_1$
Length of normal GN=CN-CG
$\Rightarrow x_1- (x_1-\large\frac{b^2}{a^2}$$x_1)$
$\Rightarrow \large\frac{b^2}{a^2}$$x_1=(1-e^2)x_1$
Hence (a) is the correct answer.
answered Feb 7, 2014 by sreemathi.v
 

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