Lines intersecting a parabola

The equation of a straight line through the origin is in the form $y=mx$ , where $m$ is the gradient. Then the two points of intersection satisfy $y = x^2 - 2$ and $y=mx$ , hence $x^2 - 2 = mx$ $\implies x^2 - mx \red{- 2} = 0$ . Let the two roots satisfying the equation be $x_1$ and $x_2$ , these are the two $x$ -coordinates of the two points. By Vieta's formula , $x_1x_2 = \red{-2}$ . Since there are four different lines, the products of all the eight $x$ -coordinates is $(-2)^4 = \boxed{16}$ .

Since the lines pass through the origin, we can assume, they're in a form $y = \alpha x$ for $\alpha \in \mathbb{R}$ .

First, let's look if there exists such a number $\alpha$ for which the line intersects the parabola only once or not at all.

We need to look at the number of solutions of a quadratic equation with a parameter.

$x^2 - 2 = \alpha x$

$x^2 - \alpha x - 2 = 0$

By examing the determinant,

$D = b^2 - 4ac = \alpha^2 - 4 \cdot (-2) = \alpha^2 + 8$

we can conclude that it will always be greater than $0$ , meaning that for any arbitrary value of $\alpha$ , the line $y = \alpha x$ interesects the parabola in exactly two points.

Next, we'll have a look at the product of x-coordinates which are given by a pair of solutions of the equation $x^2 - 2 = \alpha x$ .

$x_{1,2} = \frac{b^2 \pm \sqrt{ b^2 - 4ac}}{2a} = \frac{\alpha^2 \pm \sqrt{\alpha^2 + 8}}{2}$

Then, $x_1 \cdot x_2$ gives

$\frac{\alpha^2 + \sqrt{\alpha^2 + 8}}{2} \cdot \frac{\alpha^2 - \sqrt{\alpha^2 + 8}}{2} = \frac{\alpha^2 - (\alpha^2 + 8)}{4} = \frac{-8}{4} = -2$ .

Since we have 4 lines, we just multiply $-2$ by itself $4$ times or, equivalently, evaluate $(-2)^4 = 16$ .

Problem idea is taken from a section for 3rd and 4th year high school students in a Czech mathematical olympiad Matematický klokan . The solutions, however, are not published with procedures - this one is mine. Link to the problem statement is here .