+0

# Parabolas

0
92
2

A line $$y = mx + c$$ intersects the parabola $$y=x^2$$ at points $$A$$ and $$B$$.

The line $$AB$$ intersects the y-axis at the point $$P$$

If $$AP-BP=1$$ then find $$m^2$$. There are just so many factors to pay attention to, and I don't see any way to use the fact that $$AP-BP=1$$ without the distance formula and very complicated square roots.  To start though, I listed point A as $$(a, a^2) because(y=x^2)$$ and B as ($$b, b^2)$$, with $$a^2=ma+c$$ and $$b^2 = mb+c.$$  Point P is also (0, c).

Thanks for any help!  From a minor exploit in the "formating tips" section though, this answer should be a fraction, and contain a square root (example: $$\frac{\sqrt2}{5}$$​)

Jul 13, 2021

#1
0

If you knew the solution, can you please share it?

Jul 14, 2021
edited by Guest  Jul 14, 2021
#2
0

As to how $$m=a+b$$ :

Finding the slope between AB:
$$m=\frac{a^2-b^2}{a-b}=\frac{(a-b)(a+b)}{(a-b)}=a+b$$

$$x^2-mx-c=0$$ -->subsituite m=a+b then let x=a or x=b and solve for c, yielding c=-ab

Guest Jul 14, 2021