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If $x+y=k$ is a normal to the parabola $y^2=12 x$, then $k$ is
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The correct answer is:
$9$
Any normal is $y+t x=6 t+3 t^3$. It is identical with $x+y=k$ if $\frac{t}{1}=\frac{1}{1}=\frac{6 t+3 t^3}{k}$ $\therefore t=1$ and $1=\frac{6+3}{k} \Rightarrow k=9$.
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