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

If the line $x-1=0$ is the directrix of parabola $y^2-kx+8=0$ then one of the value of k is

$\begin{array}{1 1}(A)\;\frac{1}{8} \\(B)\;8 \\(C)\;4 \\(D)\;\frac{1}{4} \end{array}$

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1 Answer

The parabola is $y^2-Kx+8=0$
$y^2=kx-8$
$\qquad= k(x-\large\frac{8}{k})$
$\qquad=4AX$
where $4A=K$
direction of the parabola is $X+A=0$
Comparing it with $x-1=0$ we get,
$1= \large\frac{8}{k}-\frac{K}{4}$
or $k^2+4k-32=0$
$(k+8)(k-4)=0$
$k=-8,4$
Hence C is the correct answer.
answered Apr 10, 2014 by meena.p
 
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