Finding All Possible Quadratic Equations From Two Points

I'm trying to find all of the possible quadratic equations that can be found within a certain scenario. In this scenario, there are two static Cartesian points, and then there is a

Solution 1:

The general form of the parabolic functions (with a vertical axis) is

f(x)= ax² + bx +c

You impose that the points (x₁,y₁) and (x₂,y₂) must belong to the graph of the function.

That is,

y₁ = ax₁² + bx₁ +c
y₂ = ax₂² + bx₂ +c

From these we obtain

c = y₁ - ax₁² - bx₁
y₂ = ax₂² + bx₂ + (y₁ - ax₁² - bx₁) = a(x₂²-x₁²) + b(x₂-x₁) + y₁

With that restrictions, we can get rid of the parameters b and c:

    y₂ - y₁ - a(x₂² - x₁²)   y₂-y₁
b = ────────────────────── = ───── - a(x₁+x₂)
           x₂ - x₁           x₂-x₁

                y₂-y₁                       y₂-y₁
c = y₁ - ax₁² - ─────x₁ + a(x₁+x₂)x₁ = y₁ - ─────x₁ + ax₁x₂
                x₂-x₁                       x₂-x₁

So we have

             ┌ y₂-y₁            ┐                y₂-y₁
f(x) = ax² + │ ───── - a(x₁+x₂) │x + y₁ - ax₁² - ─────x₁ + a(x₁+x₂)x₁
             └ x₂-x₁            ┘                x₂-x₁ 

Simplifying a little,

       ┌          y₂-y₁ ┐                
f(x) = │a(x-x₂) + ───── │(x-x₁) + y₁
       └          x₂-x₁ ┘                

Varying a you get all the possible functions.

Solution 2:

You're on the right track. You're last equation gives you c in terms of a. Your second to last equation gives you b in terms of a. Substituting both of those into your general equation gives you all three coefficients in terms of a. Then a is the dynamic integer value you want to shape your curve.