Parabolas and hyperbolas - how do you find them?

The_ArtofScience

Finding linear equations for initial conditions has never been a problem from me since is easy to see (x1, y1) and (x2, y2) can be plugged into the slope equation. But what if I'm looking for a parabolic or even a hyperbolic line, how do I then find its equation with the initial condition that it must have some (x,y) and (x2, y2) points?

mathman

The general form for conic sections (parabolas, hyperbolas, ellipses, circles, two lines) looks like ax2+bxy+cy2+mx+ny+k=0. Since you can divide by any constant, there are five parameters. Therefore you need to specify the curve at five points.

The_ArtofScience

Well, given that I still don't see a way to make sense out of that general equation since conic secs don't have an actual slope (unless one were to find its many tangents). I'm convinced that there is a way to solve for a parabolic equation given two sets of coordinates though

maze

You plug the coordinates into the equation for each of the 5 points. That gives you 5 equations and 5 unknowns, which you can solve for.

Eg: suppose the curve passes through the following points (im just making these points up):
(1,2); (0,0); (3,4); (-1,3); (-2,-3).

Plugging each point into the equation
$$x^2+b \cdot x \cdot y+c \cdot y^2+m \cdot x+n \cdot y+k=0$$,

you get the following 5 equations:
$$1^2+b \cdot 1 \cdot 2+c \cdot 2^2+m \cdot 1+n \cdot 2+k=0$$
$$0^2+b \cdot 0 \cdot 0+c \cdot 0^2+m \cdot 0+n \cdot 0+k=0$$
$$3^2+b \cdot 3 \cdot 4+c \cdot 4^2+m \cdot 3+n \cdot 4+k=0$$
$$(-1)^2+b \cdot (-1) \cdot 3+c \cdot 3^2+m \cdot (-1)+n \cdot 3+k=0$$
$$(-2)^2+b \cdot (-2) \cdot (-3)+c \cdot (-3)^2+m \cdot (-2)+n \cdot (-3)+k=0$$

Simplifying,
$$1+2 b+4 c+m +2 n+k=0$$
$$k=0$$
$$9+12 b+16 c+3 m+4 n+k=0$$
$$1-3 b +9 c-1 m+3 n+k=0$$
$$4+6 b+9 c-2 m-3 n+k=0$$

Now you can solve these equations for b,c,m,n, and k.

----

I went ahead and solved this system of equations since it isn't much trouble, and got the following answers:
b = 116/33
c = -95/33
m = -133/11
n = 257/33
k = 0

So,
$$x^2+\frac{116}{33}x \cdot y-\frac{95}{33} y^2-\frac{133}{11} x+\frac{257}{33} y=0$$

Plotting the resulting figure, it looks like a hyperbola, and it passes through all the points we want. Whew!
http://img520.imageshack.us/img520/8662/hyperbolapl5.png [Broken]

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The_ArtofScience

Sorry that you had to go through all that trouble maze. Thanks for the example

HallsofIvy

Homework Helper
Well, given that I still don't see a way to make sense out of that general equation since conic secs don't have an actual slope (unless one were to find its many tangents). I'm convinced that there is a way to solve for a parabolic equation given two sets of coordinates though
Then you are convinced wrong. y= x2 and y= 2x2- 1
both pass through the two points (1, 1) and (-1, 1).

Even given the information that the graph is a parabola, concave upward, with vertical axis, the two points (1, 1) and (-1, 1) are not enough to define a specific parabola.

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