# Rotating 1/x to make a hyperbola?

Hey everyone, I was having trouble with this question.

The graph of 1/x is a hyperbola, but it's equation does not fit the form (x-h)/(a^2) - (y-k)/(b^2) = 1. Rotate 1/x using polar coordinates, change it back into cartesian coordinates, and write the equation in standard hyberbola notation.

I understand how to convert to polar from cartesian and vice versa, but I don't know how to rotate graphs. Any help would be great! Thanks!

Ambitwistor
When you have the expression in polar form, replace the angle &theta; with &theta;+&phi; to rotate by an angle &phi; (or &theta;-&phi;, depending on your sign conventions). You'll probably want to use the trig addition identities (for, say, sin(a+b) in terms of sines and cosines of a and b) to put your expression into a form that can be easily converted back to Cartesian coordinates.

Homework Helper
If (x,y) are the original coordinates and (x', y') are the coordinates rotated at angle &theta; then

x'= x cos(&theta;)+ y sin(&theta;);

y'= -x sin(&theta;)+ y cos(&theta;);

Notice that if you solve the two equations for x, y in terms of x',y', this is, inverting the change, you get

x= x' cos(&theta;)- y' sin(&theta;)

y= x' sin(&theta;)+ y' cos(&theta;)

Exactly what you would get if you replace &theta; by -&theta;

In particular, if &theta;= &pi;/2, then