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. 

Solutions To Analytic Geometry Problems

The Problems - Solutions:

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

Solns.

The general analytic equation is:

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


For ellipse:    Let A = 9,  B= 0, C = 4,  D= 36,  E =  -8 and F = 4

Then we get:  9 x ²  +   4  y ²   + 36 x   - 8 y + 4 = 0

With appropriate algebraic manipulation we get:

(x + 2) ² / 4 +    (y - 1 ) ²   / 9 = 1

Graphing yields:



For hyperbola: Let A = -1, B= 0,  C = 4,  D = E = F = 0

Then we get:

4  y 2   - x ²  + 4 x  = 0

Graphing yields:



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

Solution:  Take the sample parabola:  y ²  =   4 x

The general analytic equation is:

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

Then for this case: B= 0 (no xy term), and  A= 0, C = 1

Therefore:  B²   -  4AC    =   - 4 AC = - 4(0)(1) = 0

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

Solution: Use the example ellipse of Problem (1):

9 x ²  +   4  y ²   + 36 x   - 8 y + 4 = 0

B= 0,  A = 9,  C = 4

Therefore:  B²   -  4AC    =   - 4 AC = - 4(9)(4) =  -324

So the condition is satisfied

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

Solution: Using the hyperbola from Problem 1:

4  y²   - x ²  + 4 x  = 0

Therefore: B= 0,  A = -1, C = 4

Then:  B²   -  4AC    =   - 4 (-1)(4)  =  16

Which is greater than zero so the condition holds.


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.

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, after some algebra:

3(x') ²  +  ( y') ²    =  2

Divide through by 2:

3(x') ² / 2   +  ( y') ²  / 2  =  1

This curve defines an ellipse which is shown below:



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

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

A = 1, B = 1, C = 1

Then:  B²   -  4AC    =  (1)²   - 4 (1)(1)  =  1   - 4   =  -3

Since B²   -  4AC    <  0  the curve is an ellipse

b) x²  -    y²    =  1

A = 1, B = 0, C = -1

Then:  B²   -  4AC    =     - 4 (1)(-1)  =  4

Since B²   -  4AC    >  0  the curve is a hyperbola


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

A = 1, B = -3, C =  3

Then:  B²   -  4AC    =  (-3)²   - 4(1) (3) =   9  -  12 = -3

Since B²   -  4AC    <  0  the curve is an ellipse

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

A = 1,  B = 0, C = 1

Then:  B²   -  4AC    =     - 4 (1)(1)  =  - 4

Since B²   -  4AC    <  0  the curve is an ellipse

Looking At Analytic Geometry (2): The Ellipse

The ellipse is a geometric figure - curve we examined before in the context of real planet orbits. For example, in the diagram shown above, a planet at point P(x,y) in its orbit is referenced to the two foci of the ellipse at f1 and f2. As the point P moves along the total connecting distance f1-P-f2 remains constant. By inspection what we call the semi-major axis is a = 4 and the semi-minor axis is approximately 2.8.

If we know the general form of the equation for an ellipse, e.g.

x² / a ² +    y²  / b ² = 1

We could make a fair guess at what the actual equation of the orbit shown would be, say taking

b =   Ö 8  =   2.82

The ellipse equation would then be:

x² /16 +    y²   / 8 = 1

And the graph would be plotted as shown below:

Which is a solid match for the earlier sketch orbit.

What is c, mathematically?

It is the square root of the difference between the semi-major axis squared, and the semi-minor axis squared:

c =   Ö (a ²  -   b ²)

In this case, we'd get:    Ö (16  -  8) =   Ö 8  =   2.82

That means the coordinates of the foci are f1 (- Ö 8, 0)  and f2 (0,  - Ö 8)

Note that when c = 0 we have a = b. (One constant fixed radius: e.g. r = a = b)

This means the figure is a perfect circle. For a circle the eccentricity e = 0. For our own example here: e =  Ö 8  / 4   =   2.82/ 4  = 0.705

When, however, a > b then the shape alters to elliptical, but also when b > a.

If we juxtapose the values for a, b in the general equation we get:

x² / 8  +    y²   / 16 = 1

And the ellipse orientation changes to one with major axis now in the vertical direction:

From the preceding it is possible to sketch the graph of a given ellipse, as well as give its eccentricity e (= c/a) from just the coordinates of the eclipse center,. the focus F (at least one) and the semi-major axis a.

Example problem: Sketch the graph and find e for the ellipse with center coordinates C(0,0), focus coordinate F(0, 2) and a = 4

Solution: Given the coordinate F(0,2) we must have a > b   and hence, c = 2. Then the semi-minor axis is found from:

 b =   Ö (a ²  -   c ²)    =   Ö (16  -   4 )    =  Ö 12

The eccentricity e = c / a    = 2/ 4   = 0.5

The equation for this ellipse will be:

x² /16 +    y²   / 12 = 1

As in the case of the circle, ellipses aren't always centered at (0,0) or the origin of an x-y grid. Consider the equation below:

9 x²  +   4  y²   + 36 x   - 8 y + 4 = 0

To get this into a form that can easily be inspected and graphed, first re-arrange the terms to get:

9(x²  +   4 x)  +  4 (y²   - 2 y)  =  - 4

Complete the square for each term in parenthesis:

9 (x²  +   4 x  +  4)  +  4 (y²   - 2 y   + 1)  =  - 4   + 4   + 36

Or:

9 (x²  +   4 x  +  4)  +  4 (y²   - 2 y   + 1)  =  36

Divide through by 36:

(x + 2) ² / 4 +    (y - 1 ) ²   / 9 = 1

To simplify further, let x' = (x + 2) and y' = ( y - 1)

Then write:

x' ² / 4 +    y ' ²   / 9 = 1

Which yields the ellipse shown below once the x, y- coordinate offsets are taken into account:



We also find: 

c =   Ö (a ²  -   b ²)     where in this case,  a ²  =  9   and   b ²  =   4

So:  c =   Ö (9   -   4 )    =   Ö 5

So the foci are situated at the points (0,  + Ö 5 ) in the new coordinate system or at:

(-2,  1 +   Ö 5) in the original. Which values can be checked from the graph.

Problems:

1) In the last example problem worked above, we saw that the ellipse equation of form:

(x + 2) ² / 4 +    (y - 1 ) ²   / 9 = 1

Could be simplified to: x' ² / 4 +    y ' ²   / 9 = 1

Using:   x' = (x + 2) and y' = ( y - 1)

Show how the offset affects the location of the simplified ellipse form by graphing both ellipses on the same axes. Thus, satisfy yourself that the graph shown is justified once the x, y- coordinate offsets are taken into account.

2) An ellipse has center coordinates at C(-3,0) and focus at F(-3, -2) with a = 4. Find the eccentricity e and sketch the graph.

3) An ellipse has center coordinates at C(2, 2) and focus at F(-1, 2) with a = Ö 10. Find the eccentricity e and sketch the graph.

4) Sketch the following ellipses from their equations below, and give the semi-major axes, semi-minor axis and eccentricity e.

a) 4 x²  +   9  y²   = 144

b) 4 x²  +    y²   =    1

5) Look up the definition of the "directrix" of an ellipse. Then find the eccentricity and directrices of the ellipse:

x² /7 +    y²   / 16 = 1

And sketch the graph showing the locations of the foci.

Seen 'Hidden Figures'? Here's A Look At Analytic Geometry (1)

In the new movie 'Hidden Figures', about three African-American women working at NASA who are "human computers" (computing spacecraft trajectories) there occurs a compelling moment  when Kevin Costner's character ("Al Harrison"- head of NASA's Langley computing section) asks if there isn't anyone who can do analytic geometry. At that point the white female supervisor (played by Kristen Dunst) points to Katharine G.  Johnson - who also chirps up "I can!"  This familiarity serves her well in computing orbits of satellites, manned craft as well as changing orbits - say from circular or elliptical to parabolic.

Given the currency of the film, it's relevant  to look at some examples from analytic geometry and offer up some problems too, What is analytic geometry? It is basically that area of mathematics that applies algebra and calculus to geometry in order to work out the locus of points for various figures, always expressible in the form F (x, y) = 0.

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. All of these things we will do and I will challenge readers to do in assorted problems!

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 I invite readers to try. You will end up with the figure shown below:



This is actually an equation of basic form:  x ²    +    y²  =      r² 

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    +    y²   + 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  +  y²   - 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 generic analytical expression, e.g.

Ax ²    +  Ay²   + Dx  + Ey  + F   = 0   (A not equal 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 for the energetic reader.

When you do this you should obtain the general, analytical expression for the radius:

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

Let's check to see that this works out for the example problem we set, i.e.

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.

Problems for the math-minded (or readers looking for something to challenge them!)

1)  Complete the square for the general analytic equation:

Ax ²    +  Ay²   + Dx  + Ey  + F   = 0

To show how the radius r is derived, as well as the coordinates for the center of a circle not necessarily at (0,0)

2) Check the equation:  x ²    +    y²  =  4

to see that the general form also applies here.

3) Find the coordinates of the center of each of the following circles and the radius r, also sketch the circle:

a)  x ²    +    y²  - 2 y   =  3

b) 2x ²    +  2  y²   + x  + y = 0

c) x ²    +    y²  + 2x  = 8

4) Obtain the equation for the circle shown below: