Parabolas and hyperbolas - how do you find them?

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

Science Advisor
7,629
378
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.
 
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
 
654
3
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
[tex]x^2+b \cdot x \cdot y+c \cdot y^2+m \cdot x+n \cdot y+k=0[/tex],

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

Simplifying,
[tex]1+2 b+4 c+m +2 n+k=0[/tex]
[tex]k=0[/tex]
[tex]9+12 b+16 c+3 m+4 n+k=0[/tex]
[tex]1-3 b +9 c-1 m+3 n+k=0[/tex]
[tex]4+6 b+9 c-2 m-3 n+k=0[/tex]

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,
[tex]x^2+\frac{116}{33}x \cdot y-\frac{95}{33} y^2-\frac{133}{11} x+\frac{257}{33} y=0[/tex]

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]
 
Last edited by a moderator:
Sorry that you had to go through all that trouble maze. Thanks for the example
 

HallsofIvy

Science Advisor
Homework Helper
41,641
837
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.
 

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top