Thread: Curve were all points lie just as far from P as from m.

1. Curve were all points lie just as far from P as from m.

Hi, I am stuck on this task;

A curve (c) goes so that all points lie just as far from P as from m.
P(1,3) and M(x)= 2x-4

What type of curve is c?

2. Re: Curve were all points lie just as far from P as from m.

this is a parabola by definition

3. Re: Curve were all points lie just as far from P as from m.

Let $(x_0, y_0)$ be a point on this curve. Its distance from (1, 3) is $\sqrt{(x_0- 1)^2+ (y_0- 3)^2}$. Its distance from y= 2x- 4 is $\frac{2x_0- y_0- 4}{\sqrt{5}}$. Those two distances will be the same if and only if $\sqrt{(x_0- 1)^2+ (y_0- 3)^2}= \frac{2x_0- y_0- 4}{\sqrt{5}}$. Squaring both sides, $(x_0- 1)^2+ (y_0- 3)^2= \frac{2x_0- y_0- 4)^2}{5}$. Multiplying everything, $5x_0^2- 10x_0+ 5+ 5y_0^2- 30y_0+ 45= 4x_0^2+ y_0^2+ 16- 4x_0y_0- 16x_0+ 8y_0$.

$x_0^2+ 4x_0y_0+ 4y_0^2+ 6x_0- 38y_0+ 29= 0$.

The second degree terms are of the form "$Ax^2+ Bxy+ Cy^2$ with A= 1, B= 4, and C= 4. It's "discriminant" is $B^2- 4AC= 16- 16= 0$ which tells us this conic section is a parabola.

As Idea said, one of the definitions of a parabola is "a curve that is always equi-distant from a point (the "focus") and a line (the "directrix").