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(D) Let the equation of the tangent to the parabola ${y^2} = 8x$ be $y = mx + \frac{2}{m}$.Since this line is also a tangent to the hyperbola $xy = -1

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$y = x + 2$

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(D) Let the equation of the tangent to the parabola ${y^2} = 8x$ be $y = mx + \frac{2}{m}$.Since this line is also a tangent to the hyperbola $xy = -1$,we substitute $x = \frac{-1}{y}$ into the tangent equation:$y = m(\frac{-1}{y}) + \frac{2}{m}$$y^2 = -m + \frac{2y}{m}$$my^2 - 2y + m^2 = 0$.For the line to be a tangent,the discriminant of this quadratic equation must be zero:$D = (-2)^2 - 4(m)(m^2) = 0$$4 - 4m^3 = 0$$m^3 = 1 \implies m = 1$.Substituting $m = 1$ into the tangent equation $y = mx + \frac{2}{m}$,we get:$y = (1)x + \frac{2}{1}$$y = x + 2$.

The equation of the common tangent to the curves ${y^2} = 8x$ and $xy = -1$ is

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