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For a general conic

$$ ax^2 + bxy + cy^2 + dx + ey + h = 0$$

we see that, when rearranged and thought of as a quadratic in $x$, the discriminant of the resulting quadratic is

$$\Delta_x(y) = (b^2 - 4ac)y^2 + (2bd - 4ae)y + (d^2 - 4ah) = \delta y^2 + Ry + S.$$

It is easy to show that when $\delta $ is $0$ and $R > 0$, the conic is a parabola. However, the next problem has me stumped.

Show that if $\delta < 0$, $\{y: \Delta_x(y) \geq 0\}$ is one of:

i. $\emptyset$

ii. a single point

iii. a closed interval $[\alpha,\beta]$

I'd like a small hint to help me.

Soham Chowdhury
  • 1,797

1 Answers1

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I think I have a solution now. I just hit upon it by thinking of the equality somewhat geometrically: $\Delta_x(y) \geq 0$ only at the $y$-values of the points where the parabola defined by $x = \Delta_x(y)$ lies to the right of the $y$-axis.

Since we are told that $\delta < 0$, this parabola opens to the left (as $y\to \infty$, the value of $x$ goes to $-\infty$).

Hence either the number of points is $0$ (if the parabola is wholly in the open left half-plane) or $1$ (if it touches the y-axis at one point), or they all lie in some interval on the y-axis.

Soham Chowdhury
  • 1,797