## Sunday, June 26, 2011

### Different views, same hyperbola

In the algebra study guide that my son is using (ALEKS) they set up a hyperbola in two "standard" forms as:

 `(x-h)2/a2 - (y-k)2/b2 = 1(y-k)2/b2 - (x-h)2/a2 = 1`

or more generally:

 `(x-h)2/a2 - (y-k)2/b2 = +/- 1`

Let's simplify our lives and consider an example centered at the origin:

 `x2/a2 - y2/b2 = 1`

Then they divide the world into those hyperbolas that open up and down versus those that open up left and right.

By looking at the equation, I think we can agree that if x2 equals 0 we've got a problem, since no value of y can satisfy the equation, and in fact x must not be less than 1. Hence, we can already predict that this system opens left and right, as the plot shows (`a2 = b2 = 2`):

The ALEKS review goes on to explain that for this type of equation, the vertices of the hyperbola (points of closest approach to the origin at `h,k`) are equal to

 `h +/- a, k`

and the asymptotes have slopes equal to

 `+/- b/a`

This all works great. The problem I had was that the simplest parabola I can think of is:

 `xy = 1`

which doesn't fit the system.

The answer to my confusion is that any parabola may be rotated around its origin in the xy-plane. In 2 selected orientations its equation will have only `x2` and `y2` terms (when the vertices are on the x or the y-axis), whereas in 2 other selected orientations it may have only `xy` terms (when the vertices are on `y= +/- x`. The rest of the time it contains both.

This is explained in an excerpt from Stewart's Calculus I found here.

Consider a point `P` in the plane with coordinates `x,y` and distance from the origin `r`. Now, let's establish a new coordinate system `u,v` which is rotated counter-clockwise by the angle `θ`. A vector from the origin through `P` is rotated an angle `φ` with respect to the u axis and `φ + θ` with respect to the x-axis. Here's a screenshot from the pdf:

(Note, he uses capital X and Y for the second set of coordinates).

In the `u,v` system, the point `P` has coordinates

 `u = r cos(φ)v = r sin(φ)`

In the `x,y` system the coordinates are:

 `x = r cos(φ + θ)y = r sin(φ + θ)`

We remember (my post here) that:

 `sin(s+t) = sin s cos t + cos s sin tcos(s+t) = cos s cos t - sin s sin t`

Hence:

 `x = r cos(φ) cos(θ) - r sin(φ) sin(θ) = u cos(θ) - v sin(θ)y = r sin(φ) cos(θ) + r cos(φ) sin(θ) = v cos(&theta) + u sin(θ) = u sin(&theta) + v cos(θ)`

Now, consider the hyperbola:

 `xy = 1`

Rewrite this in terms of `u,v` and multiply, giving 4 terms

 `+ u cos(θ) u sin(θ)+ u cos(θ) v cos(θ)- u sin(θ) v sin(θ)- v sin(θ) v cos(θ)`

If `θ` equals `π/4`, then

 `cos(θ) = sin(θ) = 1/√2`

The two middle terms drop out and leave us with:

 `u2/2 - v2/2 = 1`

If the middle terms don't cancel, we're left with a mixture including some fraction of
u2/2, v2/2, and u times v.

Note: we rotated the coordinate system counter-clockwise, which has the effect of rotating the plot clockwise, when the coordinate system is viewed in standard orientation.