Mathematico | Mathematico

The curves

\begin{aligned} y & = p x^{2} + 2 q x + p \\ y & = 2 q + p x - q x^{2} \end{aligned}

intersect exactly once.

Find the sum of all possible values of $\frac{p}{q}$ .

Hint

The algebra gets worse before it gets better - stick with it!

Solution

Consider the equation

\begin{aligned} p x^{2} + 2 q x + p & = 2 q + p x - q x^{2} \\ (p + q) x^{2} + (2 q - p) x + p - 2 q & = 0 \end{aligned}

Exactly one intersection means exactly one solution to this equation.

Next, let's look at the discriminant, whose value must be $0$

\begin{aligned} (2 q - p)^{2} - 4 (p + q) (p - 2 q) & = 0 \\ 4 q^{2} - 4 p q + p^{2} - 4 (p^{2} - p q - 2 q^{2}) & = 0 \\ 4 q^{2} - 4 p q + p^{2} - 4 p^{2} + 4 p q + 8 q^{2} & = 0 \\ 12 q^{2} - 3 p^{2} & = 0 \\ 3 p^{2} & = 12 q^{2} \\ \frac{p^{2}}{q^{2}} & = \frac{12}{3} \\ \frac{p}{q} & = \pm 2 \end{aligned}