Common Tangent

First note that the parabola $y^{2} = 8x$ opens to the right, and the hyperbola $xy = -1$ has two branches, one in the second quadrant and one in the fourth. From a quick sketch it should be clear that the common tangent line will be tangent to the parabola in the first quadrant and tangent to the hyperbola in the fourth quadrant.

For the parabola, in the first quadrant, we have that $y = 2\sqrt{2x}$ for which $\dfrac{dy}{dx} = \dfrac{\sqrt{2}}{\sqrt{x}}$ .

For the hyperbola we have that $y = -\dfrac{1}{x}$ , for which $\dfrac{dy}{dx} = \dfrac{1}{x^{2}}$ .

Now let the points of tangency to the common tangent line be $(a, -\frac{1}{a})$ for the hyperbola and $(b, 2\sqrt{2b})$ for the parabola. We want the slopes of the respective tangent lines at these points to be equal, and so

$\dfrac{1}{a^{2}} = \dfrac{\sqrt{2}}{\sqrt{b}} \Longrightarrow b = 2a^{4} \Longrightarrow \sqrt{b} = \sqrt{2}a^{2}$ , as $b \gt 0$ .

We also require that the slope of the line joining the two points of tangency be equal to the slopes found above. This implies that

$\dfrac{2\sqrt{2b} - (-\dfrac{1}{a})}{b - a} = \dfrac{1}{a^{2}}$

$\Longrightarrow \dfrac{4a^{2} + \dfrac{1}{a}}{2a^{4} - a} = \dfrac{1}{a^2} \Longrightarrow 4a^{4} + a = 2a^{4} - a$

$\Longrightarrow 2a^{4} + 2a = 0 \Longrightarrow a(a^{3} + 1) = 0$ ,

which has the real solution $a = -1$ , as we cannot have $a = 0$ . So the point of tangency to the hyperbola is $(-1, 1)$ , yielding a slope of $\dfrac{1}{a^{2}} = 1$ . The common tangent line will then have the equation

$y - 1 = 1*(x - (-1)) \Longrightarrow \boxed{y = x + 2}$ .

Check: With $a = -1$ we have that $b = 2$ , and so the point of tangency to the parabola is $(2, 4)$ . The slope at this point is $\dfrac{\sqrt{2}}{\sqrt{b}} = 1$ as expected. Finally, we note that $(2, 4)$ does indeed lie on the line $y = x + 2$ , thus confirming that we have found the common line of tangency.

Simple algebra works does work. Let the equation be of the form y=mx+c, put this equation into the two given equations and consider their delta to be zero ( b^2=4ac ), what resulting is two equations with two unknowns, m and c, and the values for them can easily be found.

A fast, informal solution: Given the geometrical considerations already pointed out by @brian charlesworth , we expect the tangent to have an equation of the form $y=mx+q$ with $m,q>0$ . Let $a$ and $b$ denote the points where the tangent line touches respectively the hyperbola and the parabola. Then (again for the considerations above) we expect $a<0$ , $b>0$ . Moreover, we must have $\frac{1}{a^2} = m = \frac{\sqrt{2}}{\sqrt{b}}$ then equating the left and right sides we see that $a$ and $b$ must satisfy the equation $\frac{1}{a^4} = \frac{2}{b}$ , the easiest solution of which is clearly $b=2$ and $a=-1$ (rembering $a<0$ ). Then $m=1$ and find the $q$ is now straightforward.

Assume we are in $R^{2}$

Take the derivatives of each equation

$A: y^{2}=8x \Rightarrow 2y \times \frac{dy}{dx}=8 \Rightarrow \frac{dy}{dx} = \frac{4}{y} \\ B: xy=-1 \Rightarrow y=\frac{-1}{x} \Rightarrow \frac{dy}{dx}= \frac{1}{x^{2}}$

Let $P_A(\frac{a^{2}}{8}, a)$ and $P_B(b, \frac{-1}{b})$ form the common tangent be on A, B respectively.

$\therefore \frac{dy}{dx}= \frac{4}{a} = \frac{1}{b^{2}} \Rightarrow a = 4b^{2}$

By slope equation and the relationship above

$\frac{dy}{dx}= \frac{a - (\frac{-1}{b})}{\frac{a^{2}}{8} - b}=\frac{4b^{2} + (\frac{1}{b})}{2b^{4} - b} =\frac{4b^{3} + 1}{b^2(2b^{3} - 1)} = \frac{1}{b^{2}} \\ \Rightarrow 4b^{3} + 1 = 2b^{3} - 1, b^{3}=-1 \Rightarrow b=-1cis(0), -1cis(2\pi/3), -1cis(4\pi/3)$

reject $b \not\in R^{2}$ to get b = -1.

Substitute b = -1 into derivative of curve B: $\frac{dy}{dx}= \frac{1}{(-1)^{2}} =1$

Equation of common tangent: $x - y = -1 - \frac{1}{1}=-2 \Rightarrow \boxed{y=x+2}$