What is the necessary and sufficient condition for a point to be an interior point of a parabola?

Question

Point M is called an interior point of the given parabola if any line passing through the M point (not parallel to the parabola axis) intersects this parabola at two different points. What is the necessary and sufficient condition for the point (X₀, Y₀) to be an interior point of a parabola $y^2 = 2px $?

Answer 1

Let $P$ be the parabola with equation $y^2=2px$.

Let $M(x_0,y_0)$ with $x_0 \neq 0$ (the summit is a particular case).

The general equation of a line passing through $M$ is

$$(L_m) : \ \ \ y=y_0+m(x-x_0)\tag{0}$$

Thus, the abscissas of the intersection points of line $L_m$ with parabola $P$ are such that :

$$(y_0+m(x-x_0))^2=2px$$

an equation that we can reorder as a quadratic in variable $x$:

$$ m^2x^2 + 2(- m^2x_0 + my_0 - p)x+ (mx_0 - y_0)^2 = 0 $$

with discriminant

$$\Delta=4(- m^2x_0 + my_0 - p)^2-4(m(mx_0 - y_0))^2 $$

Taking into account the form $4(A^2-B^2)=4(A+B)(A-B)$ of $\Delta$, we get the simplified form :

$$\Delta=4p(2m^2x_0+2my_0+p)\tag{1}$$

Now, taking your definition, point $M$ is interior to parabola $P$ if, whatever $m$, there are two (real !) roots in (1), which is expressed by condition :

$$\forall m, \ \ \Delta>0\tag{2}$$

Due to the fact that the coefficient of $m^2$ in (1) is $>0$, condition (2) is ensured if equation

$$\Delta=0 \ \iff \ 2x_0m^2+2y_0m+p=0,\tag{3}$$

considered as a quadratic in $m$, has no real roots.

This is equivalent to the fact that the discriminant $\delta$ of (3) :

$$\delta=4y_0^2-8px_0=4(y_0^2-2px_0)$$

is negative, whence the necessary and sufficient condition :

$$y_0^2<2px_0$$

of "insideness" as could be awaited.