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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Coordinate Geometry

The normal at $(a,2a)$ on $y^2=4ax$ meets the curve again at $(at^2,2at)$ then the value of t is

$\begin{array}{1 1}(A)\;1 \\(B)\;3 \\(C)\;-1 \\(D)\;-3 \end{array}$

1 Answer

If the normal at $(at^2,2at)$ on $y^2=4ax$ meets the curve again at $(at_2^2,2at_2)$ then $t_2=-t-\large\frac{2}{t_1}$.
The value of parameter $t_1$ for the point $(a,2a)$ is given by
$at_1^2=a$
$2at_1=2a$
=> $t_1=1$
=> $t_2=-1-(2/-1)$
=> $\quad= -3$
Hence $t=-3$
Hence d is the correct answer.
answered Apr 9, 2014 by meena.p
 

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