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# Find the point of intersection of tangents at any two points on the parabola $y^2=4ax$?

$\begin{array}{1 1}(a)\;x=at_1t_2\\(b)\;x=2at_1t_2\\(c)\;x=a(t_1+t_2)\\(d)\;x^2=a(t_1t_2)\end{array}$

+1 vote
Equation of tangent at $(at_1^2,2at_1)$ is
$t_1y=x+at_1^2$-------(1)
Equation of tangent at $(at_2^2,2at_2)$ is
$t_2y=x+at_2^2$--------(2)
(1) -(2) we get,
$(t_1-t_2)y=a(t_1-t_2)(t_1+t_2)$
$y=a(t_1+t_2)$
$x=t_1a(t_1+t_2)-at_1^2$
$x=at_1t_2$
Hence (a) is the correct answer.