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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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If $x+y=k$ is the normal to $y^2=12x$ then k is

$\begin{array}{1 1}(A)\;3 \\(B)\;9 \\(C)\;-9 \\(D)\;-3 \end{array}$

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Comparing $y^2=12x$ with standard equation of parabola $y^2=4ax$ we have $a=3$
$y=mx+c$ is a normal to parabola if $C= -2am-am^3$
Comparing $x+y=k$ with $y=mx+c$ we get,
$m=-1 \qquad c=k$
$k=-2a(-1)-a(-1)^3$
$k= -6(-1)-3(-1)$
$\quad= 6+3$
$\quad=9$
Hence B is the correct answer.
answered Apr 10, 2014 by meena.p
 

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