Solving for a and b easily: (p-a)^2 / ((b-a) (c-a)) = r, (b-q)^2 / ((b-a) (b-c)) = r

[qwertytam]

Hi, I'm trying to solve the two equations below for a and b, with the constants c, p, q, and r known. Also, a < p < c < q < b.

(p-a)^2 / ((b-a) (c-a)) = r
(b-q)^2 / ((b-a) (b-c)) = r

If I multiply these equations out, it starts getting ugly real quick. I can see it is possible to get to the point where you can use the quadratic equation to solve, but if I do that, the number of terms quickly multiplies. Is there a way to elegantly arrive at two equations to solve for a and b?

Thank you

 

[qwertytam]

Yes, so I can simplify to: (p-a)^2 / (c-a) = (b-q)^2 / (b-c)

I can then rearrange to a quadratic equation in terms of a: a^2 + a(B-2p) + (p^2-Bc) = 0, where B = (b-q)^2/(b-c)

That then gives me a solution for a:

a = (B-2p) +/- SQRT((B-2p)^2 - 4(p^2-Bc)) / 2

I then get stuck, because if I try and solve the equation for a in terms of b, well it is just ugly with 4th powers etc. Am I missing a step to make the solution simpler?

As a reminder, where I'm trying to get to is a solution that gives me values for a and b, given the constants c, p and q.

 

[qwertytam]

So I see that by simplifying we can go from: \(\displaystyle \frac{(p-a)^2}{(b-a)(c-a)} = \frac{(b-q)^2}{(b-a)(b-c)}\)

To: \(\displaystyle \frac{(p-a)^2}{(c-a)} = \frac{(b-q)^2}{(b-c)}\)

If I let: \(\displaystyle B = \frac{(b-q)^2}{(b-c)}\)

Then: \(\displaystyle p^2 - 2pa + a^2 = Bc - Ba\)

I can then form a quadratic equation: \(\displaystyle a^2 + a(B-2p) + p^2-Bc = 0\)

Then, using the quadratic formula, I get a solution for a: \(\displaystyle a = \frac{(2p-B) \pm \sqrt{(B-2p)^2 - 4(p^2-Bc)}}{2}\)

At this point, I stop for a few reasons:
1) To solve the quadratic formula, I assume I would need to substitute back in for B, which would make this very ugly to solve
2) Substituting the algebraic solution for a into one of the previous equations also seems like a sure route into algebraic ****

Am I missing a (hopefully) simpler path to find a solution for both a and b?