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If $x-2 y+k=0$ is a tangent to the parabola $\mathrm{y}^2-4 \mathrm{x}-4 \mathrm{y}+8=0$, then the value of $\mathrm{K}$ is
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Verified Answer
The correct answer is:
$7$
$y=\mathrm{m} x+\mathrm{c} \Rightarrow x-2 y+k=0$
$$
\begin{aligned}
& m=\frac{1}{2}, c=\frac{k}{2} \\
& y^2-4 x-4 y+8=0 \\
& (y-2)^2=4(x-1)
\end{aligned}
$$
From (i), $x=2 y-k$
$$
\begin{aligned}
& (y-2)^2=4(2 y-k-1) \\
& y^2-12 y+(8-4 k)=0
\end{aligned}
$$
$\because$ line is tangent
$\therefore$ Above quadratic equation must have equal roots i.e. $\mathrm{D}=0$
$$
(12)^2-4(8-4 k)=0 \Rightarrow k=7
$$
$$
\begin{aligned}
& m=\frac{1}{2}, c=\frac{k}{2} \\
& y^2-4 x-4 y+8=0 \\
& (y-2)^2=4(x-1)
\end{aligned}
$$
From (i), $x=2 y-k$
$$
\begin{aligned}
& (y-2)^2=4(2 y-k-1) \\
& y^2-12 y+(8-4 k)=0
\end{aligned}
$$
$\because$ line is tangent
$\therefore$ Above quadratic equation must have equal roots i.e. $\mathrm{D}=0$
$$
(12)^2-4(8-4 k)=0 \Rightarrow k=7
$$
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