Question

If the line $2x-1=0$ is the directrix of the parabola $y^2-kx+6=0$, then one of the values of $k$ is

• A. -6
• B. 6
• C. 1/4
• D. -1/4

Answer 1

Answer: (A)

Given $e{q}^n$ of parabola is
$y^2-kx+6=0$
$\Rightarrow y^2=kx-6$
$\Rightarrow y^2=k\left(x-\frac{6}{k}\right)$
Now, directrix, $x-\frac{6}{k}=-\frac{k}{4}$
$\Rightarrow x=\frac{6}{k}-\frac{k}{4}...(i)$
But directrix is given
$\Rightarrow x=\frac{1}{2}...(ii)$
$\therefore \frac{6}{k}-\frac{k}{4}=\frac{1}{2}$
$\Rightarrow 24-k^3=2k$
$\Rightarrow k^2+2k-24=0$
$\Rightarrow (k+6)(k-4)=0$
$\Rightarrow k=-6,k=4$