Thread: How to Determine a Hyperbola From Points

Hi,
Is there a way to determine a hyperbola (equation) from a few points (three maybe)?

I thought that there is only one way, which is by graphing and finding the vertex, transverse axes, etc... but I hope there are other ways to determine and equation.

Thanks!

Originally Posted by masoug

Hi,
Is there a way to determine a hyperbola (equation) from a few points (three maybe)?

I thought that there is only one way, which is by graphing and finding the vertex, transverse axes, etc... but I hope there are other ways to determine and equation.

Thanks!

If you have the general equation of a hyperbola a set of points defines a set of simultaneous equations in the parameters which can then be solved.

CB

Originally Posted by CaptainBlack

If you have the general equation of a hyperbola a set of points defines a set of simultaneous equations in the parameters which can then be solved.

CB

Um, can you explain what you mean by that?

If you know that a hyperbolas axes are parallel to the x and y axes, then a hyperbola can be written as either

or


That depends on 4 parameters, , , a, and b. If you know 4 points you can put the x,y values of those points into the equation, getting 4 equations to solve for the parameters.

If it is possible that the axes of the hyperbola are tilted with respect to the x and y axes the problem becomes much more complicated. However, any conic section (hyperbola, parabola, ellipse, circle and some special cases) can be written in the form . That has 6 parameters but we could always divide the entire equation by one of the so there are 5 independent parameters. That means that 5 points are sufficient to determine any conic section.

Originally Posted by HallsofIvy

If it is possible that the axes of the hyperbola are tilted with respect to the x and y axes the problem becomes much more complicated. However, any conic section (hyperbola, parabola, ellipse, circle and some special cases) can be written in the form . That has 6 parameters but we could always divide the entire equation by one of the so there are 5 independent parameters. That means that 5 points are sufficient to determine any conic section.

I see...
So for most conic sections, just use the relation and pick six points to satisfy each unknown.

Thanks!

For all conic sections and 5 points will be sufficient.