Common normal for an ellipse and parabola

Question

I came across the following question in a magazine with example problems for the JEE examination in India:

Given are two curves $x^2/a^2 + y^2/63=1$ and $y^2=4x$.

The maximum integral value of $a$ for which there is only one common normal to the two curves is:

A. 7
B. 8
C. 9
D. 10

My attempts so far:

1. I came up with an equation for the normal of an ellipse in terms of slope, and one for parabola, then solved them, but the calculations got really messy.
2. Used a graph plotter, but could not digest the fact that they have a common normal.

How can this problem be solved?

Answer 1

The only parabola normal passing through $(0,0)$ or parallel to the axes is $x=0$ and it is always normal to the ellipse. Thus, we can parametrize every other potential normal as $$ \frac xp+\frac yq = 1. $$

Together with normal for parabola equation ($2y/q=4/p$) at point $x,y$, we get: $$ p=2+x,\qquad q=\frac{2+x}2y\\ \text{and since $y^2=4x$:}\qquad q^2 = p^2(p-2) $$

Doing the same thing with ellipse ($2x/a^2p+2y/63q=0$), we get: $$ p=x\left(1-\frac{63}{a^2}\right),\qquad q=y\left(1-\frac{a^2}{63}\right),\\ \text{and since $x^2/a^2+y^2/63=1$:}\qquad a^2p^2+63q^2=(a^2-63)^2 $$

Now we can formulate the question as

What is the largest parameter $a$ such that curves (1) $q^2=p^2(p-2)$ and (2) $a^2p^2+63q^2=(a^2-63)^2$ do not intersect$^*$?

One can show that $p_1=2$ is the minimal $p$ value of (1) and $p_2=|a-63/a|$ is the maximum $p$ value of (2). Solving $p_1=p_2$ gives you two values of $a=7$, $a=9$. So if $a\in[7,9]$, then there is only one common normal.

${}^*$ intersection where $p\neq0$, since $p,q\neq 0$ from the equation of normal