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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 hyperbola)

$\begin{array}{1 1}(a)\;x_1-\large\frac{a^2}{x_1},\normalsize (e^2-1)x_1\\(b)\;x_1-\large\frac{a}{x_1},\normalsize (e^3-1)x_1\\(c)\;x_1^2+\large\frac{a^2}{x_1},\normalsize (e-1)x_1\\(d)\;x_1+\large\frac{a^2}{x_1},\normalsize (e^2+1)x_1\end{array}$

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1 Answer

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Let the tangent and normal at $4P(x_1,y_1)$ meet the x-axis at $T$ and $C$
Equation of tangent at $P(x_1,y_1)$
$\large\frac{xx_1}{a^2}-\frac{yy_1}{b^2}$$=1$------(1)
T lies on x-axis ,put $y=0$ in (1)
$\Rightarrow x=CT$
$CT=\large\frac{a^2}{x_1}$
$CN=x_1$
Length of subtangent NT=CN-CT
$\Rightarrow x_1-\large\frac{a^2}{x_1}$
Equation of normal at $P(x_1,y_1)$ is
$\large\frac{a^2x}{x_1}+\frac{b^2y}{y_1}$$=a^2+b^2$-----(2)
G lies on x-axis ,y=0 in (2)
$x=CG$
$CG=\large\frac{(a^2+b^2)x_1}{a^2}$
Length of sub-normal NG=CG-CN
$\Rightarrow \large\frac{(a^2+b^2)x_1}{a^2}$$-x_1$
$\Rightarrow \large\frac{b^2x_1}{a^2}$
$\Rightarrow (e^2-1)x_1$
Hence (a) is the correct answer.
answered Feb 10, 2014 by sreemathi.v
edited Mar 21, 2014 by balaji.thirumalai
 

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