## Thursday, February 16, 2017

### Analytic Geometry Conclusion: Generalizing 2nd Degree Curves While we have examined separately the circle, ellipse, parabola and hyperbola it should also be noted that all are second degree curves - what we call "conic sections" - which can be portrayed as special cases of a general analytic equation, viz.

Ax2  +    Bxy +   Cy2   + Dx  + Ey  + F   = 0

For example, the circle template equation with center at (h, k)  e.g.

(x - h)2    + (y - k ) 2   =  r2

Can be extracted from the general analytic equation by taking:

A = C = 1     B = 0,   D = -2h

E = - 2k,  F =  h2    +  k 2   -    r2

The parabola specific equation:   x 2  =    4py

Is obtained by using:

A = 1,  E = - 4p, and B = C = D = F = 0

Note that the terms:   Ax2 ,      Bxy ,     Cy2

are the second degree or quadratic terms with  Bxy the "cross product" term

Most problems to do with this term involve eliminating the cross product using a rotation of Cartesian axes such that:

(x)
(y)  =

(cos Θ.   -    sin  Θ) (x')
(sin Θ.    +  cos Θ)  .(y')

Performing the matrix operation for rotation:

x = x' cos Θ   -    y' sin  Θ

y = x' sin Θ.   +  y'  cos Θ

What if  Θ  =  45 degrees?

Then  cos (45)  = sin (45)    =  1 / Ö 2

Then:  x = x' / Ö 2   -   y' / Ö 2  =    (x'   -   y') / Ö 2

y  = x' / Ö 2   -+  y' / Ö 2  =   ( x'   +   y' ) / Ö 2

Now, if  we apply the rotation of axes equation to the general analytic equation for 2nd degree curves what do we get? It can be shown we have:

A'x' 2  +    B' x' y' +   C' y' 2   + D' x'  + E'y'  + F'    = 0

With the following relations to the original coefficients:

A'  =   A cos2 Θ  +   B cos Θ sin  Θ  + C sin 2 Θ

B'    = B (cos2 Θ  - sin 2 Θ) + 2 (C - A) sin  Θ cos Θ

C =   A sin 2 Θ  -   B  sin  Θ cos Θ +   C cos2 Θ

D'   =  D cos Θ + E sin  Θ

E' =   - D  sin  Θ   + E'  cos Θ

F'    = F

The technique for getting rid of the cross product terms is pretty straightforward, given an angle or rotation  Θ  can always be found such that the new cross product term is eliminated.  How to do this? Simply set  B' = 0 in the 2nd equation in the set above and solve for the angle Θ. Of course, it helps to have at hand a couple of trig identities:

cos2 Θ  -    sin 2 Θ    =    sin 2Θ

And:

2  sin  Θ cos Θ      = sin 2Θ

So that, for example:

B' =  B cos 2Θ  +   (C - A) sin 2Θ

So B' will vanish if we choose:

cot 2Θ  =   (A  - C)/ B

Example Problem: Eliminate the cross product term in the equation:

x2  +   xy  +  y2    =  3

And thereby identify the 2nd degree curve and graph  it.

Solution:

The given equation has: A = B = C = 1

Choose Θ  according to: cot 2Θ  =   (A  - C)/ B  =   (1 - 1)/ 1 = 0/1 = 0

Then:  2Θ  =  90 deg     so: Θ  =  45 deg

Then:  x  =   (x'   -   y')  / Ö 2

y = (x'   +   y' ) / Ö 2

And:

3(x') 2  +  ( y') 2    =  6

Divide through by 6:

(x') 2 / 2   +  ( y') 2  / 6  =  1

which is an ellipse with foci on the y' - axis. This is sketched below: Problems:

1)  Use the general analytic equation to write specific equations for the ellipse and hyperbola. Sketch the resulting curves to confirm your answers.

2) The discriminant:  B2   -  4AC     =   B'2   -  4A'C'

has been found valid for any rotation of axes (i.e. for any angle Θ )  Thus, the quantity is invariant under a rotation of axes.

a) Show that for the parabola the discriminant:  B2   -  4AC    =  0

b) Show that for the ellipse:   B2   -  4AC    <  0

c) Show that for the hyperbola:   B2   -  4AC    >  0

3) Determine the equation to which:

x2  +   xy  +  y2    =  1

reduces when the axes are rotated to eliminate the cross product term. Sketch the resulting curve.

4) Use the discriminant to classify each of the following curves as a circle, parabola, ellipse, or parabola:

a) x2  +   y2    +   xy  + x  -   y    =  3

b) x2  -    y2    =  1

c)   x2  +  3 y2    -  3 xy   -  6 y    =  7

d) x2  +   y2    +  3 x  -  2 y    =  3