The fourth point

Geometry Level 4

Consider all parabolas of the form $y={{x}^{2}}+{{a}^{2}}x+2a$ , for $a \in \mathbb{R}$ , which intersect the coordinate axes in three distinct points. For such $a$ , denote by $C_{a}$ the circle through these three intersection points. Every circle $C_{a}$ has one more common point with the parabola. The locus of this fourth point is $\text{\_\_\_\_\_\_\_\_\_\_} .$

Bonus : Find the exact locus of the point of our interest.

Inspiration

part of a parabola two half-lines a circle a parabola a straight line an arc of a circle a half-line

1 solution

David Vreken
Nov 23, 2020

The $y$ -intercept of $y = x^2 + a^2x + 2a$ is $(0, 2a)$ , and the $x$ -intercepts are $(\frac{1}{2}(-a^2 - \sqrt{a^4 - 8a}), 0)$ and $(\frac{1}{2}(-a^2 + \sqrt{a^4 - 8a}), 0)$ .

The center of the circle will lie on the vertical line through the midpoint of $(\frac{1}{2}(-a^2 - \sqrt{a^4 - 8a}), 0)$ and $(\frac{1}{2}(-a^2 + \sqrt{a^4 - 8a}), 0)$ , which is $x =-\frac{1}{2}a^2$ .

Since the parabola has a vertical axis, the fourth point will be the reflection of the $y$ -intercept in the line $x =-\frac{1}{2}a^2$ , so that $x = -a^2$ and $y = 2a$ . After eliminating $a$ , these two equations solve to $y^2 = -4x$ , which is a parabola.

However, since there are no $x$ -intercepts when $a^4 - 8a < 0$ , which is when $0 < a < 2$ , some of the parabola is skipped. Therefore, the locus of the fourth point is part of a parabola .

The fourth point

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