Math Question - Lines of tangency to two parabolas...

Factual Questions
1

So my entire calculus class was stumped by a problem today and no one seemed to get anywhere… The problem gives two parabolas and asks for the equations for two lines that are tangent to both parabolas… The problem states:

Sketch the graphs of the two equations y=x^2 and y=-x^2+6x-5, sketch two lines that are tangent to both graphs. Find the equations of the lines.

We seemingly tried just about everything for 45 minutes to no avail. The back of the book reveals the answers as y=2x-1 and y=4x-4, yet we still have no clue to get to those answers… Anyone here know of the method?

2

Do you have a graphic calculator?

3

You can calculate the formulas for tangent line at a given x value for each of the parabolas, using the knowledge that the derivative is the slope of the tangent line, and the line has to pass through the function value:

y = 2mx - m^2, line tangent to f(x) = x^2 at m
y = (6-2n)x + n^2 - 5, line tangent to f(x) = -x^2 + 6x - 5 at n

For the lines to be the same, the slope and intercepts must be the same:

2m = 6 - 2n
-m^2 = n^2 - 5

substitute m or n from the first equation into the second, and you have a quadratic to solve for the appropriate values for one of the equations.

4

I just worked it out using brute force, and I do agree with those answers.

First off, can you verify for yourself that the answers given in the back are, in fact, correct? Make sure that you know what you’re looking for in a solution. :slight_smile:

(I see on preview that yabob has answered - more elegantly - but I’ll plunge ahead nevertheless.)

For any one parabola, you can obtain a relationship between the slope and y-intercepts of all lines which come tangent to it at some point.

Consider the parabola y = x[sup]2[/sup]. Note that y’ = 2x.

A tangent line must, at some x, match both y and y’. A line will be of the form y = mx + b, so:

mx + b = x[sup]2[/sup] (match y-value)
m = 2x (match slope)

Substitute out the x, and you get b = -(m[sup]2[/sup]/4). Any line for which m and b relate like that comes tangent to the first parabola.

Now, do the same thing for the second parabola. The math is a bit more complicated, but it’s the same idea. You’ll there get (in this case) another quadratic relating m and b for tangent lines to that parabola.

Any (m,b) pairs that satisfy both relationships represent lines that come tangent to both parabolas. You have two equations for the two unknowns. There are quadratic terms, so there are two solutions. (m,b) = (4,-4) and (m,b) = (2,-1) work.

5

I would have done it the same way as yabob, but just for fun, I did it without calculus. I got the same answer, which is good. Here’s how I did it, if you’re interested.

These two parabolas both have focal parameter (1/2), so they’re essentially identical, but rotated 180°. So, you can use symmetry and determine that a line which is tangent to both must pass through the midpoint of the vertices of the two parabolas, (3/2, 2). So the question is equivalent to, what line passes through the point (3/2, 2) and is tangent to the parabola y = x[sup]2[/sup].

Now draw the tangent line to the parabola and notice two things, which follow from the basic properties of parabolas. First, the distance between the tangent point and the focus is the same as the distance between the tangent point and the directrix. Second, the angle formed by the line from the focus to the tangent point is the same as the angle formed by the line perpendicular to the directrix. (cf. Fig. 1)

Using vertical angles and similar triangles, we can show that the distance between (3/2, 2) and the focus is the same as the distance between (3/2, 2) and the directrix point. (cf. Fig. 2)

We know the coordinates of the focus, and we know the y-coordinate of the directrix. So we can solve for the distance that I labelled d in Fig. 2:ul[sup]2[/sup] + (7/4)[sup]2[/sup] = d[sup]2[/sup] + (9/4)[sup]2[/sup][/ul]Solving gives d = -1/2, +1/2. Thus the x-coordinate of the tangent point must be either 1 or 2.

Going with the first choice, the tangent point is at (1, 1). The equation of the line passing through both (1, 1) and (3/2, 2) is y = 2x - 1. Going with the second choice, we similarly get y = 4x - 4.

6

Another method without calculus, just to be different:

Assume the line is of the form y=mx+b. This line intersects the parabola y=x^2 at points where x^2=mx+b. If the line is tangent to the parabola, then the equation x^2=mx+b has exactly one solution, which implies that m^2+4b=0. (m^2+4b is the discriminant: the part under the square root sign in the quadratic formula.)

Then repeat with the second parabola, and continue as in brad_d’s solution.