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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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The normal to the parabola $y^2=8x$ at the point $(2,4)$ meets the parabola again at the point.

$\begin{array}{1 1}(A)\;(-18,-12) \\(B)\;(-18,12) \\(C)\;(18,12) \\(D)\;(18,-12) \end{array}$

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1 Answer

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$y^2=8x$ -----(1)
=> $ 2y. \large\frac{dy}{dx}$$=8$
Slope of tangent at $(2,4)= \bigg[\large\frac{dy}{dx}\bigg]_{(2,4)}$
Slope of normal at $(2.4)=-1$
Equation of normal at $(2,4)$ is
$x+y-6=0$ ----(2)
Solving (1) and (2) we get,
and $x=18,y=-12$
$\therefore $ other point is $(18,-12)$
Hence D is the correct answer.
answered Apr 9, 2014 by meena.p

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