multiple tangent lines

[orangecrush]

Find the equation of the line(s) that are tangent to the graphs of y=x^2 and y=-x^2 + 6x - 5. (it says I will need to calculate slopes several ways)

I tried by taking the derivative of each equation, which would be the slopes, and making them equal to each other. Dont know what that got me though.

 

[pka]

The slopes have to be equal:
\(\displaystyle \begin{array}{l}
y = x^2 \quad \Rightarrow \quad y' = 2x \\
y = - x^2 + 6x - 5\quad \Rightarrow \quad - 2x + 6 \\
\end{array}.\)
This happens when\(\displaystyle 2x = - 2x + 6\quad \Rightarrow \quad x = \frac{3}{2}.\)
What y value on each curve corresponds to \(\displaystyle x = \frac{3}{2}?\)
Now write the equation.

 

[soroban]

Hello, orangecrush!

pka, you found a value of \(\displaystyle x\) where the tangents are parallel.
This is not the location of a tangent common to both curves.


I found a solution . . . but there must be a simpler way.

Find the equation of the line(s) that are tangent to the graphs of \(\displaystyle y\:=\:x^2\) and \(\displaystyle y\:=\:-x^2\,+\,6x\,-\,5\)
(It says I will need to calculate slopes several ways)

Let \(\displaystyle P(p,\,p^2)\) be a point on \(\displaystyle y\,=\,x^2\)
The derivative is: \(\displaystyle \,y'\,=\,2x\)
The slope at \(\displaystyle x\,=\,p\) is: \(\displaystyle \,2p\)

Let \(\displaystyle Q(q,\,-q^2+6x-5)\) be a point on \(\displaystyle y \,=\,-x^2\,+\,6x\,-\,5\)
The derivative is: \(\displaystyle \,y;\,=\,-2x\,+\,6\)
The slope at \(\displaystyle x\,=\,q\) is: \(\displaystyle \,-2q\,+\,6\)

We want the slopes to be equal, so we have: \(\displaystyle \,2p\:=\:-2q\,+\,6\;\;\Rightarrow\;\;p \:=\:3\,-\,q\;\) [1]

The slope of \(\displaystyle PQ\) is: \(\displaystyle \L\:\frac{p^2\,-\,(-q^2\,+\,6q\,-\,5)}{p\,-\,q} \;= \;\frac{p^2\,+\,q^2\,-\,6q\,+\,5}{p\,-\,q}\)

This must equal the slope at \(\displaystyle P\), so we have:
. . \(\displaystyle \L\frac{p^2\,+\,q^2\,-\,6q\,+\,5}{p\,-\,q}\)\(\displaystyle \:=\:2p\;\;\Rightarrow\;\;p^2\,+\,q^2\,-\,2p\,-\,4q\,+\,5\:=\:0\)

Substitute [1]: \(\displaystyle \3\,-\,q)^2\,+\,q^2\,-\,2(3\,-\,q)\,-\,4q\,+\,5\:=\:0\)

. . which simplifies to: \(\displaystyle \:q^2\,-\,4q\,+\,4\:=\:0\;\;\Rightarrow\;\;(q\,-\,2)^2\:=\:0\;\;\Rightarrow\;\;q\,=\,2\)

Hence: \(\displaystyle \,p\,=\,1,\;q\,=\,2\) . . . the slope is \(\displaystyle m\,=\,2\)
. . and the points are: \(\displaystyle \,P(1,1),\;Q(2,3)\)

The line through \(\displaystyle (1,1)\) with slope \(\displaystyle m\,=\,2\) is:
. . \(\displaystyle y\,-\,1\:=\:2(x\,-\,1)\;\;\Rightarrow\;\;\L y\:=\:2x\,-\,1\)

 

[orangecrush]

thanks for that explanation. i'm sure that has to be the easiest way to do that.
I put the tangent line and the two graphs on a graphing calculator and it seems there is another possible tangent line. would you know how to get that other tangent line?

 

[pka]

The other is y=4x-4.

 

[orangecrush]

How did you get (p^2 + q^2 - 6q + 5)/(p - q) = 2p.... to be p^2 + q^2 - 2p - 4q +5 = 0. ?
I tried simplifying it myself, and I got p^2 + q^2 - 6q + 5 = (2p)(p - q) which does not simplify to what you got above. what did you do?