The equation of the parabola is y^2 = 4x.

The equation of the hyperbola is xy = 2.

To find the equation of a common tangent to both curves, we need to find the point of tangency.

Let's first find the point of tangency on the parabola.

Since the parabola is symmetric about the y-axis, the point of tangency will have the form (a, 2√a) or (a, -2√a).

Substituting the coordinates into the equation of the parabola, we get:

(2√a)^2 = 4a

Simplifying, we get:

4a = 4a

So, any point of the form (a, 2√a) or (a, -2√a) lies on the parabola.

Next, let's find the point of tangency on the hyperbola.

To do this, we'll differentiate the equation of the hyperbola implicitly with respect to x:

y + xy' = 0

Simplifying, we get:

y' = -y/x

We want the tangent line to be perpendicular to the tangent line on the parabola, which means the slopes of the two lines should be negative reciprocals of each other.

So, the slope of the tangent line to the hyperbola at the point (a, 2√a) or (a, -2√a) should be -1/(-2√a/a) = 1/√a.

Let's call the slope of the tangent line to the hyperbola m.

m = 1/√a

Now, we can use the point-slope form of the equation of a line to find the equation of the tangent line:

y - 2√a = (1/√a)(x - a)

Simplifying, we get:

√a(y - 2√a) = x - a

Taking the square of both sides to eliminate the square root, we get:

a(y^2 - 4a) = (x - a)^2

Expanding and rearranging, we get:

x^2 - 2ax + a^2 - ay^2 + 4a^2 = 0

Since we want the equation of the common tangent, we can eliminate the parameter a:

x^2 - (y^2 - 4x) = 0

x^2 - y^2 + 4x = 0

So, the equation of a common tangent to the parabola y^2 = 4x and the hyperbola xy = 2 is x^2 - y^2 + 4x = 0.