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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Find the equation of the chord of the parabola $y^2=12x$ which is bisected at $(5,-7)$

$\begin{array}{1 1}(A)\;6x-7y+19=0\\(B)\;6x+7y+19=0 \\(C)\;6x+9y+20=0 \\(D)\;3x+4y+12=0 \end{array}$

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1 Answer

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Here, $x_1=5;\qquad y_1=-7 ;\qquad a=3$
Equation of the chord is $S_1=T$
$(-7)^2-12(5)=y(-7)-6(x+5)$
=> $6x+7y+19=0$
Hence B is the correct answer.
answered Apr 9, 2014 by meena.p
 

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