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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Find the point of intersection of normals at any two points on the parabola $y^2=4ax$

$\begin{array}{1 1}(a)\;k=-at_1t_2(t_1+t_2)\\(b)\;k=-at_1t_2(t_1-t_2)\\(c)\;k=-at_1t_2(t_2-t_1)\\(d)\;\text{None of these}\end{array}$

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Let point of intersection S(h,k).
Let two points at which normal is drawn be $(at_1^2,2at_1)$ & $(at_2^2,2at_2)$
Equation of normal at $t_1$ is
$y=-t_1x+2at_1+at_1^3$--------(1)
Equation of normal at $t_2$ is
$y=-t_2x+2at_2+at_2^3$--------(2)
(2)-(1) we get,
$x=2a+a(t_1^2+t_2^2+t_1t_2)$
$y=-at_1t_2(t_1+t_2)$
Hence $h=2a+a(t_1^2+t_2^2+t_1t_2)$
$k=-at_1t_2(t_1+t_2)$
Hence (a) is the correct answer.
answered Feb 6, 2014 by sreemathi.v
 

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