Solving Tangent Lines for 2 Graphs: x1,x2

[Michele Nunes]

Homework Statement

Sketch the graphs of y=x² and y= -x²+6x-5, and sketch the two lines that are tangent to both graphs. Find equations of these lines.

Homework Equations

The Attempt at a Solution

So I know that a tangent line to both graphs means that the tangent line will touch the first graph at (x₁,y₁) and the second graph at (x₂, y₂) and at these separate points, the derivatives will be equal. So my first equation looks like 2x₁ = -2x₂+6, however I'm stuck on what other equation I'm supposed to use to solve for both x-variables.

 

[BvU]

You have the sketch, I suppose ?
A line is $y = ax + b$; its intersections with the two parabolas follow from two quadratic equations that both have to have only one solution.

 

[Michele Nunes]

You have the sketch, I suppose ?
A line is $y = ax + b$; its intersections with the two parabolas follow from two quadratic equations that both have to have only one solution.

I'm sorry but I still don't understand how that relates to what I'm supposed to do. They're supposed to have only one solution to what? If I set ax+b equal to the quadratic equations, there's still too many variables for me to solve for which is where I'm confused.

 

[LCKurtz]

Maybe this will help. Let $P=(a,a^2)$ be an unknown point on the parabola $y=x^2$. Write the equation of the tangent line at that $P$. Of course, it will depend on the unknown $a$, but that's OK.
Now look at a general point on the second parabola: if $x=b$ then $y = -b^2+6b-5$, so the general point on that parabola is $Q=(b,-b^2+6b-5)$. Now what needs to happen for the tangent line to the first parabola at $P$ to be tangent to the second parabola at $Q$? That should give you some equations in $a$ and $b$ to work with. Sorry, I changed your notation a bit to avoid all the subscripts.

 

[Michele Nunes]

Maybe this will help. Let $P=(a,a^2)$ be an unknown point on the parabola $y=x^2$. Write the equation of the tangent line at that $P$. Of course, it will depend on the unknown $a$, but that's OK.
Now look at a general point on the second parabola: if $x=b$ then $y = -b^2+6b-5$, so the general point on that parabola is $Q=(b,-b^2+6b-5)$. Now what needs to happen for the tangent line to the first parabola at $P$ to be tangent to the second parabola at $Q$? That should give you some equations in $a$ and $b$ to work with. Sorry, I changed your notation a bit to avoid all the subscripts.

The derivatives have to be equal at P and Q, right? So I set the derivatives equal, but I need another equation in order to solve for both variables. Is there something I can do with the original equations? I can't set them equal though because the points don't happen at the same y-value.

 

[LCKurtz]

The tangent line at $P$ must pass through $Q$.

 

[Michele Nunes]

The tangent line at $P$ must pass through $Q$.

I finally got it, I've been working on this problem for so long. Thank you so much!