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(D) Let the equation of the tangent to the parabola $y^2=4x$ be $y=mx+\frac{1}{m}$.For the parabola $x^2=-32y$,the equation of the tangent with slope

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$x-2y+4=0$

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(D) Let the equation of the tangent to the parabola $y^2=4x$ be $y=mx+\frac{1}{m}$.For the parabola $x^2=-32y$,the equation of the tangent with slope $m$ is $y=mx-a m^2$,where $x^2=4ay$ gives $a=-8$.Thus,the tangent is $y=mx-(-8)m^2$,which simplifies to $y=mx+8m^2$.Comparing the two tangent equations,we have $\frac{1}{m}=8m^2$.This implies $m^3=\frac{1}{8}$,so $m=\frac{1}{2}$.Substituting $m=\frac{1}{2}$ into the first tangent equation: $y=\frac{1}{2}x+\frac{1}{1/2} = \frac{1}{2}x+2$.Multiplying by $2$,we get $2y=x+4$,or $x-2y+4=0$.

The equation of the line touching both parabolas $y^2=4x$ and $x^2=-32y$ is

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