Find the equation of the common tangents Find the equation of the common tangents to the parabola $y^2=4ax$ and $x^2=4by$. 
My Attempt :
$y^2=4ax$ is the equation of parabola with focus at $(a,0)$ 
$x^2=4by$ is the equation of parabola with focus at $(0,b)$
The equation of the tangent to the parabola $y^2=4ax$ is 
$$y=mx+\dfrac {a}{m}$$
$$m^2x-my+a=0$$
The equation of the tangent to the parabola $x^2=4by$ is:
$$x=m'y+\dfrac {b}{m'}$$
$$(m')^{2}y - m'x+b=0$$
 A: Let $A(at^2,2at)$ and $B(2bu,bu^2)$ be the points on $C_1:y^2=4ax$ and $C_2:x^2=4by$ respectively.
Equation of tangent of $C_1$ at $A$:
$$\frac{x}{-at^2}+\frac{y}{at}=1 \quad \cdots \cdots \: (1)$$
Equation of tangent of $C_2$ at $B$:
$$\frac{x}{bu}+\frac{y}{-bu^2}=1 \quad \cdots \cdots \: (2)$$
Comparing $(1)$ and $(2)$,
$$
\left \{
  \begin{array}{rcl}
    -at^2 &=& bu \\
    at &=& -bu^2 \\
  \end{array}
\right.$$
On solving,
$$t=\frac{1}{u}=-\sqrt[3]{\frac{b}{a}}$$
The common tangent is
$$-\frac{x}{\sqrt[3]{ab^2}}-\frac{y}{\sqrt[3]{a^2b}}=1$$
$$\fbox{$\sqrt[3]{a}\, x+\sqrt[3]{b} \, y+\sqrt[3]{a^2 b^2}=0$}$$
A: The equation of the tangent to the parabola  $y^2=4ax$ is
$$m^2x-my+a=0$$
this equation also satisfies $x^2=4by$ because you want a common tangent thus it touches the 2nd parabola at some point 
$$m^2x-m\frac{x^2}{4b}+a=0$$
find x put it back in  $x^2=4by$
you will get a point P follow the same procedure for the 2nd line get point Q  then find the line joining P AND Q
A: You are on the right tack! According to your work
$mx-y+\dfrac {a}{m}=0$ and  $x-m'y-\dfrac {b}{m'}=0$ are equations of the same line which implies that their coefficients are proportional (we assume that $a$ and $b$ are not zero):
$$\frac{m}{1}=\frac{-1}{-m'}=\frac{\dfrac {a}{m}}{-\dfrac {b}{m'}}.$$
Hence,
$$\begin{cases}
m=\dfrac{1}{m'}\\
m=-\dfrac{am'}{bm}
\end{cases}\implies 
m=-\frac{a}{bm^2}\implies m=-\sqrt[3]{\frac{a}{b}}
$$
and, by after solving with respect to $m$, we find that the the common tangent is 
$$-\sqrt[3]{\frac{a}{b}}\, x-y-\dfrac{a}{\sqrt[3]{\frac{a}{b}}}=0$$
that is
$$y=-\sqrt[3]{\frac{a}{b}}\, x-\sqrt[3]{a^2b}.$$
A: You’re almost there. You’re looking for the common tangents—the two tangent equations that you’ve written down must describe the same lines. Assuming both parabolas are nondegenerate, multiply the first equation by $b$ and the second by $a$ to get $$m^2bx-mby+ab=0 \\ m'^2 ay - m'ax + ab =0.$$ Equate coefficients to produce the system $$m^2b = -m'a \\ m'^2a = -mb$$ from which $$m^4(b/a) = -m$$ We know that the common tangent isn’t horizontal or vertical because the only place where that could happen is at the origin, but the two parabolas intersect there. Therefore, $m = -\root3\of{a/b}$.
