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

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Aicharya · Accepted Answer

Given equation 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) $\begin{aligned} & \therefore \frac{6}{k}-\frac{k}{4}=\frac{1}{2} \Rightarrow 24-k^2=2 k \ & \Rightarrow k^2+2 k-24=0 \ & \Rightarrow(k+6)(k-4)=0 \quad \Rightarrow k=-6, k=4 \end{aligned}$

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

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