[Math] Maximum number of common normal of $y^2=4ax$ and $x^2 = 4by$

Question

Maximum number of common normal of $y^2=4ax$ and $x^2 = 4by$

Attempt: Equation of normal to the curve $y^2=4ax$ in slope form $y=mx-2am-am^3.$

Now above equation is also normal to $x^2=4by$

I could not understand how to solve further, thanks.

Answer 1

In general, a normal to $y^2=4ax$ has the form

$$tx+y-2at-at^3=0$$

and a normal to $x^2=4by$ has the form

$$x+sy-2bs-bs^3=0$$

A common normal of the two parabolae has its equation expressible in both the above two forms. Obviously, $s\ne0$ and $t\ne0$.

The second equation can be written as

$$tx+sty-2bst-bs^3t=0$$

Therefore, $st=1$ and

\begin{align*} -2at-at^3&=-2bst-bs^3t\\ &=-2b-bs^2\\ &=-2b-\frac{b}{t^2}\\ 2at^3+at^5&=2bt^2+b \end{align*}

Let $f(t)=at^5+2at^3-2bt^2-b$. Then $f(-t)=-at^5-2at^3-2bt^2-b$.

If $a$ and $b$ are of the same sign, then the coefficients of $f(t)$ has one sign change and the coefficients of $f(-t)$ has no sign change. So $f(t)$ has one positive root and no negative root.

If $a$ and $b$ are of opposite signs, then the coefficients of $f(t)$ has no sign change and the coefficients of $f(-t)$ has one sign change. So $f(t)$ has no positive root and one negative root.

Therefore, $f(t)$ has only one real root. The two parabolae have at most one common normal.