Rotating 1/x to make a hyperbola?

[Khan]

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!

 

[HallsofIvy]

If (x,y) are the original coordinates and (x', y') are the coordinates rotated at angle θ then

x'= x cos(θ)+ y sin(θ);

y'= -x sin(θ)+ y cos(θ);

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(θ)- y' sin(θ)

y= x' sin(θ)+ y' cos(θ)

Exactly what you would get if you replace θ by -θ

In particular, if θ= π/2, then

x= (√(2)/2) (x'- y'); y= (√(2)/2)(x'+ y')

xy= (1/2)(x'²- y'²)= x'²/2- y'²/2= 1.