Question

How do you find the equation of a parabola when given two points (2,10) and (4,10)?

Answer 1

There's not enough information here to uniquely determine the parabola, but enough to find:

`y = ax^2-6ax+(8a+10)` for some constant `a in RR`

Explanation:

Since the `y` coordinate of both points is the same, the axis must be midway between the two points at `x=3`.

So the equation of the parabola takes the form:

`y = a(x-3)^2+k`

If we substitute `x = 4` and `y = 10` into this equation we get:

`10 = a(4-3)^2 + k = a+k`

So `k = 10-a` and we can write the equation as:

`y = a(x-3)^2+(10-a)`

`=ax^2-6ax+9a+10-a`

`=ax^2-6ax+(8a+10)`

The constant `a` can be given any value in `RR`, except that when `a=0` the 'parabola' is the straight line `y=10`.

Here are the parabolas for `a=+-1` and `a=+-2`

graph{(y-x^2+6x-18)(y+x^2-6x-2)(y-2x^2+12x-26)(y+2x^2-12x+6)=0 [-2.585, 7.415, 7.22, 12.22]}