# Finding two points...

• Sep 24th 2009, 05:35 PM
jllinas
Finding two points...
Ive been trying to figure out this problem for a while now... I can solve it graphically but I am trying to solve it algebraically using the derivative but I am still stuck:
On the curve y=x^3, find two points, P and Q, for which the tangent line at P crosses Q. Furthermore, the slope of the tangent line at Q is 4 times the slope of the tangent line at P.
I got this far then I hit a roadblock:

q = 2p.... the two tangent lines:

3p^2(x-p) + p^3 = 12p^2(x-2p) + 8p^3
• Sep 25th 2009, 04:45 AM
Opalg
Quote:

Originally Posted by jllinas
Ive been trying to figure out this problem for a while now... I can solve it graphically but I am trying to solve it algebraically using the derivative but I am still stuck:
On the curve y=x^3, find two points, P and Q, for which the tangent line at P crosses Q. Furthermore, the slope of the tangent line at Q is 4 times the slope of the tangent line at P.
I got this far then I hit a roadblock:

q = 2p.... the two tangent lines:

3p^2(x-p) + p^3 = 12p^2(x-2p) + 8p^3
If P is the point $(p,p^3)$ then the tangent at P has slope $3p^2$ and equation $y=3p^2x-2p^3$. It meets the curve again at Q = $(q,q^3)$, where q=–2p (note the minus sign). The slope at Q is $3q^2 = 3(-2p)^2 = 12p^2$. So the slope of the tangent line at Q is always 4 times the slope of the tangent line at P, no matter where P may be on the curve.