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(D) The equation of any tangent to the parabola $y^{2}=4x$ is given by $y=mx+\frac{1}{m}$.This line also touches the parabola $x^{2}=-32y$. Substituti

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$x-2y+4=0$

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(D) The equation of any tangent to the parabola $y^{2}=4x$ is given by $y=mx+\frac{1}{m}$.This line also touches the parabola $x^{2}=-32y$. Substituting $y=mx+\frac{1}{m}$ into the second equation:$x^{2}=-32(mx+\frac{1}{m})$$x^{2}+32mx+\frac{32}{m}=0$Since the line is a tangent,the discriminant of this quadratic equation must be zero:$D = (32m)^{2}-4(1)(\frac{32}{m}) = 0$$1024m^{2}-\frac{128}{m}=0$$1024m^{3}=128$$m^{3}=\frac{128}{1024}=\frac{1}{8}$$m=\frac{1}{2}$Substituting $m=\frac{1}{2}$ into the tangent equation $y=mx+\frac{1}{m}$:$y=\frac{1}{2}x+\frac{1}{1/2}$$y=\frac{1}{2}x+2$$2y=x+4$$x-2y+4=0$

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

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