+0  
 
0
40
2
avatar+179 

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
avatar
0

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

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

6 Online Users

avatar