Expert: Steve Holleran - 4/15/2007

Question
consider the three points (0,0), (1,1) and (2,4)
1. determine the equation of a parabola that passes through these three points with a vertical axis of symmetry and with a horizontal axis of symmetry
using (y=ax^2+bx+c)
2. determine the equation of a circle that passes through these three points
3. determine the intersection of the two parabolas found in question one
4. does the point found in question (3) likes on the circle found in question (2)?  why?


Answer
Hi Ken,

Wow, there sure is a lot to do here!!

Okay, let's see what we can do.

1. Using the y = ax^2 + bx + c form for the vertical parabola,

(0,0) --->  0 = c    so now y = ax^2 + bx

(1,1)--->   1 = a + b

(2,4)--->   4 = 4a + b

Solving the last two simultaneously gives a = 1 and b = 0,
so this is the standard parabola y = x^2.

For the horizontal parabola, using x = ay^2 + by + c, we have :
(0,0)---->  0 = c    so now x = ay^2 + by

(1,1)---->  1 = a + b

(2,4)--->   2 = 16a + 4b  and solving these simultaneously
gives a = -1/6, b = 7/6  so the equation is

    x = -1/6 y^2 + 7/6

2.  For the circle, the equation is

  (x-h)^2 + (y-k)^2 = r^2    so now

(0,0)--->  h^2 + k^2 = r^2

(1,1)--->  (1-h)^2 + (1-k)^2 = r^2

(2,4)--->  (2-h)^2 + (4-k)^2 = r^2

Now there's probably an easier way to do this, but what I did was :

Take the first two equations, and set them equal to each other, since they're all equal to r^2:
 h^2 + k^2 = (1-h)^2 + (1-k)^2

and if you expand the right side and simplify, you will get

  h + k = 1

Likewise if you take the third equation with the first one,

 h^2 + k^2 = (2-h)^2 + (4-k)^2

you will end up with 20 = 4h + 8k

So, solving simultaneously h + k = 1 and

                         4h + 8k = 20

you will get h = -3, k = 4 , so therefore r = 5.

The circle's equation is (x+3)^2 + (y-4)^2 = 25



For the intersection points, you already know the three that have been given:  (0,0), (1,1), (2,4).

If you take the parabola equations from part 1, you can do this:

y = x^2              and    x = -1/6 y^2 + 7/6 y

so substitute the second expression for x in the first equation:

  y = [-1/6 y^2 + 7/6 y]^2

expanding,

  y = 1/36 y^4 - 14/36 y^3 + 49/36 y^2

bring over the y on the left:

  0 = 1/36 y^4 - 14/36 Y^3 + 49/36 y^2 -y

multiply all by 36:

  0 = y^4 - 14 y^3 + 49 y^2 - 36y

factor out y:

  0 = y( y^3 - 14 y^@ + 49 y - 36)

Then I set up synthetic division on the factor in parentheses, and we know from the given points that y = 1 and y = 4 will go through, so if you do that with each of them, you will also get y = 9 at the end.

So for y = 9,  x = -1/6 (81) = 7/6 (9)

                = -81/6 + 63/6 = - 18/6 = -3.

So the intersection points for the two parabolas will be :

 (0,0),  (1,1) ,  (2,4)  ,  (-3,9).


4.  Checking (-3,9) in the circle equation, you have

  (-3 + 3)^2 + (9 - 4)^2 = 25

       0     +    25     = 25

so yes, this point is also on the circle.


Whew!!!
Thanks for giving me the workout, Ken.
It took me a little while to get this to work out.

I hope you get it in time to help you out.

Steve Holleran