Revisiting Analytic Geometry (1)

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Analytic geometry is essentially the application of algebra to geometry, whereby we are able to reference any number of two dimensional curves, e.g. circle, ellipse, parabola, and hyperbola, in terms of algebraic equations.  It is first important to see how each of these curves is formed, by taking 2 dimensional cuts of conic sections.   These curves are then recognized as distinct equations.

Once we generate the curve F(x,y) = 0,  we can then study its peculiar geometric properties (for example those for a parabola differ from those for an ellipse) as well as find the equation of a curve once its geometric properties are known

 The basic conic sections (2nd degree curves)  are shown in the diagram above. As seen there, the circle is formed from a straight cut in one plane, the ellipse is formed from a cut at an oblique angle with the plane now at some angle – say Θ  - from the horizontal.  The parabola represents an even greater oblique cut and which lacks any end closure. I.e. it is open at one end.  The hyperbola carries this further, with the cut going through the conic section on one side only – so as to form two different curves facing each other.

   Probably the most straightforward equation and curve to begin examining is the circle. Let us consider one defined by the equation:

x ²    +    y ²  =  4

This is very simple to plot as a graph and one ends up with the figure shown below:

This is actually an equation of basic form: 

x ²    +    y ²  =      r ² 

where r denotes the radius. Since the square root of  4  =  r²   is 2, then we see from the graph the radius is 2.  But this simple example is deceiving because it may give the impression that all equations of the circle consist of three terms, and all circles are centered at the origin (0, 0). In fact, most interesting cases are circles which are off -center, and also of more complicated equations.


Consider, for example, the equation:

x ²    +    y ²   + 2x   - 4y  =  11

Is this a circle? Yes, it is, but we must work to get it into the correct form to analyze, or graph.
First, write the left side as:

x ²    +  2x  +  y ²   - 4y 

Now complete the square for the expression in x and y to get:

(x + 1) ²    + (y - 2) ²   =  x ²    +  2x  + 1  +    y²   - 4y    + 4

Re-configure the entire equation paying attention to terms added to both sides:

(x ²    +  2x  + 1)   +    (y²   - 4y    + 4)   = 


11 +   4   +   1 =   16


And if we now write in the most general form using coordinates (h, k) for the center:
(x - h)²    + (y - k ) ²   =  r ² 

We see at a glance, that h = -1 and k =2 with the radius r = the square root of 16 or 4.  We can check this by doing the graph:



Inspection of the above graph shows that indeed:   (h, k) = (-1, 2), the coordinates of the center and the radius is 4 units. There is also another, longer way to approach the problem using a general analytic equation, viz.


Ax ²  +    Bxy +   Cy ²   + Dx  + Ey  + F   = 0


This can be reduced to the earlier equation:

 (x - h) ²    + (y - k ) ²   =  r ²    

By completing the square, which is left as an exercise.

The general, analytical expression for the radius is then obtained:

  r ²     = (D ²   +    E ²      - 4AF) /  4 A ²    

We check to see that this works for the example problem:

x ²    +    y²   + 2x   - 4y  =  11

Rewritten as:

x ²    +    y²   + 2x   - 4y  -  11  = 0

So:  A = 1,  D = 2,   E = -4  and F = -11

Then:

r ²     = [2 ²   +    (-4) ²    - 4 (1)(-11))]  /  4  ²  

r ²     =  [4 + 16  + 44]/   4   =   64/  4   =   16

And since:  r ²     =  16 then r = 4  which  checks out.  The coordinate of the center will also be found to be:


h =  (-D/ 2A)   and k =  (-E/ 2A)

Again, checking this for the example:

h =   -2/ 2(1) =   -1   and k  =  -(-4)/ 2 =  4/2 = 2


so (h, k ) =   (-1, 2)

which again checks.


We conclude that the circle with center at some graphical coordinate (h, k) can be expressed:

(x - h)²    + (y - 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 it 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

Example: 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

These have the following relations to the 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 trigonometric identities:

cos² Θ  -    sin ² Θ    =    sin 2Θ 

And:

2  sin  Θ cos Θ      = sin 2Θ

So that:

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

So B' will vanish if we choose:

cot 2Θ  =   (A  - C) / B

Example : Eliminate the cross product term in the equation:

x ²  +   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       

And:   y = (x'   +   y' ) / Ö 2  
  
Now write:

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.

Problems for Math Mavens:

1)Find the coordinates of the center of each for the following circles and the radius r. Sketch each of the circles

a)  x ²    +    y ²  - 2 y   =  3

b) 2x ²    +  2  y ²   + x  + y = 0

c) x ²    +    y²  + 2x  = 8

2) Use the general analytic equation to write  the specific equation for the ellipse  with:

A = 9,  B= 0, C = 4,  D= 36,  E =  -8 and F = 4

And sketch the resulting ellipse.

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.