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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Find the length of tangent,subtangent,normal and subnormal to $y^2=4ax$ at $(at^2,2at)$

$\begin{array}{1 1}(a)\;2at\sqrt{1+t^2},2a\sqrt{1+t^2},2at^2,2a\\(b)\;2at\sqrt{1-t^2},2a\sqrt{1-t^2},2at^,4a\\(c)\;4at\sqrt{1+t^2},4a\sqrt{1+t^2},2at,8a\\(d)\;2at^2\sqrt{1-t^2},2a\sqrt{1-t^2},2at,6a\end{array}$

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1 Answer

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PT=Length of tangent
PG=Length of normal
TN=length of subtangent
NG=Length of subnormal
Equation of tangent at $(at^2,2at)$
Slope of tangent $m=\large\frac{1}{t}$
Let tangent makes angle $\psi$ with +ve direction.
$t=\cot \psi$
Length of tangent at $(at^2,2at)=2atcosec \psi$
$\Rightarrow 2at\sqrt{1+\cot^2\psi}$
$\Rightarrow 2at\sqrt{1+t^2}$
Length of normal $=2at\sec\psi$
$\Rightarrow 2at\sqrt{t^2+1}$
Length of subtangent $=2at\cot\psi$
$\Rightarrow 2at^2$
Length of subnormal $=2at\tan\psi$
$\Rightarrow 2a$
Hence (a) is the correct answer.
answered Feb 6, 2014 by sreemathi.v

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