So, I have this curve C: y=2x^3 and a point P(p, 2p^3), p being a positive number. With the following graphic:
L1 and L2 are tangent lines of C passing through P.
I am asked to express the slope of L2 in terms of p.
So, I have this curve C: y=2x^3 and a point P(p, 2p^3), p being a positive number. With the following graphic:
L1 and L2 are tangent lines of C passing through P.
I am asked to express the slope of L2 in terms of p.
Let $\displaystyle m$ be the slope of $\displaystyle L_2$.
Then the equation of $\displaystyle L_2$ is: $\displaystyle y-2p^3=m(x-p)$.
$\displaystyle L_2$ must intersect the graph in two distinct points. One point is $\displaystyle P$ and the other is the point where $\displaystyle L_2$ is tangent to the graph.
So, the system $\displaystyle \left\{\begin{array}{ll}y-2p^3=m(x-p)\\y=2x^3\end{array}\right.$ must have two solutions.
Replacing y from the second equation in the first equation we have
$\displaystyle 2(x^3-p^3)=m(x-p)\Leftrightarrow (x-p)(2x^2+2px+2p^2-m)=0\Rightarrow x_1=p$ and $\displaystyle 2x^2+2px+2p^2-m=0$.
The last equation must have equal roots. Then
$\displaystyle 4p^2-16p^2+8m=0\Rightarrow m=\frac{3p^2}{2}$
Here's a calc way if you'd still like to see it.
We can find where L2 is tangent to C by:
$\displaystyle y'=6x^{2}=m$
$\displaystyle 2x^{2}-2p^{3}=6x^{2}(x-p)$
Solve this for x and we get $\displaystyle x=\frac{-p}{2}$
Sub this back into $\displaystyle y=2x^{3}$ and we see $\displaystyle y=\frac{-p^{3}}{4}$
Now, we have two end points from which to find the slope of L2. Use the slope of a line formula.
$\displaystyle m=\frac{2p^{3}+\frac{p^{3}}{4}}{p+\frac{p}{2}}=\fr ac{3p^{2}}{2}$
Which agrees with red dogs clever method.
Why yes, even shorter. I wonder why they have theta in there when it's not really needed to find the slope of L2. L1 is easy. Now that we have L2, theta can be found by:
$\displaystyle tan{\theta}=\frac{m_{2}-m_{1}}{1+m_{1}m_{2}}$, if need be.