Hyperbolas defined by quadratic equations

The following applet shows the two hyperbolas that are defined by the equations \(ax^2-cy^2=1\) (that is the red one) and \(-ax^2+cy^2=1\) (the dark violet one), respectively.

Also shown in the picture are the points with coordinates \(u = 1/\sqrt{a}\) and \(v = 1/\sqrt{c}\) on the coordinate axes (so our equations become \((\frac{x}{u})^2-(\frac{y}{v})^2 = 1\) and \(-(\frac{x}{u})^2+(\frac{y}{v})^2 = 1\), respectively), and the pair of lines solving the equation \(ax^2-cy^2=0\) (that is, \((\frac{1}{u}x+\frac{1}{v}y)(\frac{1}{u}x-\frac{1}{v}y)=0\)); these lines are the asymptotes for both of the hyperbolas.

You may adjust the values of \(a\) and \(c\) by moving the respective points in the applet with your mouse - just be sure to keep \(a\) to the right and \(c\) to the left of the black point marked \(0\). The hyperbolas and their asymptotes will be changed accordingly.

What happens if you move both \(a\) and \(c\) to the same side with respect to the black point \(0\) - although you have been warned?

In that case, one of your hyperbolas will vanish (the corresponding equation has no solutions in the real numbers any more). The other hyperbola becomes an ellipse. In fact, you have changed the signs of \(a\) and \(b\) so that they are the same, and the equation became either \(-(\frac{x}{u})^2 - (\frac{y}{v})^2=1\) (with no solutions at all), or \((\frac{x}{u})^2 + (\frac{y}{v})^2=1\) (giving you an ellipse).

A remark on printing: If you try to print the current page directly from your browser, you will probably end up with an empty grey rectangle instead of the picture. Use the PDF-file or take a screen shot.


Created by M. Stroppel using Cinderella

Powered by MathJax