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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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The correct answer is:
-6
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)
\(\begin{aligned}
& \therefore \frac{6}{k}-\frac{k}{4}=\frac{1}{2} \Rightarrow 24-k^2=2 k \\
& \Rightarrow k^2+2 k-24=0 \\
& \Rightarrow(k+6)(k-4)=0 \quad \Rightarrow k=-6, k=4
\end{aligned}\)
\(\Rightarrow y^2=k x-6 \Rightarrow y^2=k\left(x-\frac{6}{k}\right)\)
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)
\(\begin{aligned}
& \therefore \frac{6}{k}-\frac{k}{4}=\frac{1}{2} \Rightarrow 24-k^2=2 k \\
& \Rightarrow k^2+2 k-24=0 \\
& \Rightarrow(k+6)(k-4)=0 \quad \Rightarrow k=-6, k=4
\end{aligned}\)
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