Thursday, February 16, 2017

Analytic Geometry Conclusion: Generalizing 2nd Degree Curves

Image result for conic sections image
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 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:
No automatic alt text available.

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



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