Find the equation of the common tangents

Question

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: $$x=m'y+\dfrac {b}{m'}$$ $$(m')^{2}y - m'x+b=0$$

Answer 1

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

$$-\frac{x}{\sqrt[3]{ab^2}}-\frac{y}{\sqrt[3]{a^2b}}=1$$

$$\fbox{$\sqrt[3]{a}\, x+\sqrt[3]{b} \, y+\sqrt[3]{a^2 b^2}=0$}$$

Answer 2

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

Answer 3

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 $$y=-\sqrt[3]{\frac{a}{b}}\, x-\sqrt[3]{a^2b}.$$

Answer 4

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, $m = -\root3\of{a/b}$.