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