Two parabolas of the form y = ax^2+bx are tangent to the lines y = -2x-1 and y = 4x-4. Find the distance between the vertices of the parabolas.

Answer 1

When I solved this problem I used x₁ and x₂ to represent the x-coordinates of the points of tangency. For the suggestions given here let's use P and Q instead.

Because the points of tangency (POT) must lie on the parabola(s) and on the given lines. The y coordinates of the POTs must satisfy the equations of the parabola(s) and the equations for the lines.

aP² + bP = 4P - 4 (the y coordinate determined by equation of parabola = y coordinate on the line 4x-4)

aQ² + bQ = -2Q - 1 (ditto, but for POT on -2x -1)

We can also get two equations based on the fact that the derivative gives the slope at the POT.

y' = 2ax + b

So,

2aP +b = 4

2aQ + b = -2

So now we have 4 equations and 4 unknowns( a, b, P, Q).

Solving this system is tedious and requires careful calculations.

1: aP² + bP -4P = -4

2: aQ² + bQ +2Q = -1

3: 2aP + b = 4

4: 2aQ+ b = -2.

Hints:

Eliminate b from 3 and 4.

Eliminate b from 1 and 3. (solve 3 for b and substitute into 1)

Eliminate b from 2 and 4. (similar)

A: 2aP - 2aQ = 6

B: -aP² = -4

C: -aQ² = -1

Now with 3 equations and 3 unknowns, perhaps eliminate a from two different pairs (A and B) and (A and C).

Let's see if you can take it from there. The result will show that there are two possible values for Q (for example) which yield two sets of values for (a,b, P, Q), i.e. two parabolas and two POTs for each.

Here is a link to a Desmos graph that shows the results. Of course after you have the equations of the two parabolas, you must determine the corresponding vertices and then use the distance formula to find the distance between them.

desmos.com/calculator/7mgctj6ghx

Try to complete the solution yourself before looking at the graph. If you look through the equations on Desmos you will find the equations of two parabolas (one using a and b, the other hard-coded).