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If the line $2 \mathrm{x}-3 \mathrm{y}=\mathrm{k}$ touches the parabola $\mathrm{y}^{2}=6 \mathrm{x},$ then find the value of $\mathrm{k}$
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The correct answer is:
$-27 / 4$
$$
\text { } \begin{aligned}
\text { Given } x=\frac{3 y+k}{2} & \ldots \ldots . . \\
\text { and } y^{2}=6 x & \\
\Rightarrow y^{2}=6\left(\frac{3 y+k}{2}\right) \\
\Rightarrow y^{2}=3(3 y+k) \Rightarrow y^{2}-9 y-3 k=0 & \ldots \ldots . .
\end{aligned}
$$
If line (1) touches parabola (2) then roots of quadratic equation ( 3 ) is equal $\therefore(-9)^{2}=4 \times 1 \times(-3 k) \Rightarrow k=-27 / 4$
\text { } \begin{aligned}
\text { Given } x=\frac{3 y+k}{2} & \ldots \ldots . . \\
\text { and } y^{2}=6 x & \\
\Rightarrow y^{2}=6\left(\frac{3 y+k}{2}\right) \\
\Rightarrow y^{2}=3(3 y+k) \Rightarrow y^{2}-9 y-3 k=0 & \ldots \ldots . .
\end{aligned}
$$
If line (1) touches parabola (2) then roots of quadratic equation ( 3 ) is equal $\therefore(-9)^{2}=4 \times 1 \times(-3 k) \Rightarrow k=-27 / 4$
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