Solutions To Problems For Analytic Geometry Revisited (2)

The Problems Again:

1)Given  V(-3, 1) and F (0, 1) find the equation of the associated parabola and also for its directrix. Sketch the graph showing the focus, vertex and directrix. 

2)  Find the tangent to the parabola y  =   x²     using the fact that the equation of the line tangent to a curve y = f(x) is given by:

y - y1 = f'(x1) (x - x1)

where (x1, y1) = P1 is the point of the curve for which the tangent is constructed and y1  =   (x1)²     .  Show the graph of the given parabola and the tangent line with the point P1 (x1, y1) identified.

3)Analyze the equation below using the translation of axes approach and graph the resulting curve:

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

Solutions:

1) From the vertex coordinates we have: h = -3 and k = 1

With these values this indicates a parabola opening to the right  with the general form:

(y -   k) ²  =    4p (x -  h)

applies, so:

(y -   1) ²  =    12 (x  + 3 )

For which 4p  = 12 and p = 3, so we need a directrix equation of form x = -p

The graph is shown below:

2)  We have:  f'(x1)  = dy/ dx =  2x

So:  y - y1 = 2x1  (x - x1)

But only one point satisfies y1  =   (x1)²   

That is x1 = 1, y1 = 1

Hence:

y - 1 = 2 (x -1) or:   y = 2x  - 1

And this tangent line is shown below for the point P1 (x1, y1) = (1,1)

3) Collect x and y terms separately:

(x ²   -   2 x)   -   4  (y ²   -  2 y)    =  7  

(x ²   -   2 x    +   1)  - 4(y ²   -  2 y   + 1) =  7  + 1  - 4 =  5

Or:

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

Let:   x’  =  x +  1,   and:   y’  = y  -  1

Then:   x’  ² /  4   -   y’  ² /  1   =   1

Graphing:

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. 

Analytic Geometry Conclusion: Generalizing 2nd Degree Curves

                                                    

                                    Fig. 1.Conic sections generated in analytic geometry

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:

                           
                                   Fig. 2 - Ellipse for reference

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

A Look At Analytic Geometry (3): Parabolas

From parabolic telescope mirrors to receivers, the parabola plays a major role, and so it is useful to study the properties of this particular figure. Probably the classic parabola of form:

y  =   x²   

is a good place to start. A graph of this equation is shown below:



Note the vertex is on the origin or (0,0), but as with the circle and ellipse this need not be so, and one can encounter parabolas that are displaced, e.g. away from the x-axis, as well as oriented in a different way. For example, consider:  x²     = -  y  or    y  =  -   x²    


And the graph shown below:


One can also obtain parabolas with vertices oriented to the left or right of the y-axis. It is left to readers to try to find equations that yield these.

The general form of the parabola is often given as:   x²  =  -  4py 

If the parabola opens downward as shown above. If the reverse is true it is given as: x²  =  4py  . In each case, the focus of the parabola is on the axis of symmetry (e.g. follow the y-axis through the vertex downwards - or upwards in the earlier case) and p units from the vertex.

Let's consider the parabola shown  for   x²  =  -  y    then:

- 4p =   1  and p = - 1/4

That is, -  1/4 unit from the vertex to get to the focus F. What would we have for the previous parabola? Well, you should be able to work out: p = 1/4

Now, the directrix of the parabola will be the line y = p or y =  -1/4 for the downward oriented parabola. It is a line parallel to the x-axis and by that amount of displacement.  It should therefore be no surprise that the tangent to a parabola at its vertex is parallel to the directrix.  Consider a parabola of form:

x²  =   4py 

Then the slope of the tangent at any point is dy/dx = x/ 2p. which is 0 at the origin.  But the second derivative:  d² y /  dx²   = 1/2p which is positive.(So the curve is concave upward).

Of course, not all forms for the parabola are as straightforward as shown. Consider the equation:

2 x²  +   5y   - 3x + 4 = 0

We want to try to get this into the more general form for the parabola:

(x -   h) ²  =   - 4p (y -  k)

We can begin by dividing both sides of the initial equation by 2:

 x²  +   5y /2   - 3x/2  + 2   = 0

Which can also be written as:

x²     - 3x/2  =    - 5y /2   -  2

We can then complete the square, i.e. adding (- 3/4)²   to both sides:

x²     - 3x/2   + 9/16  =    - 5y /2   -  2   + 9/16

Or, write as:

(x  - 3/4)²     =   -5y/ 2  - 23/16

We can factor out 5/2 from the right hand side, i.e.:

(x  - 3/4)²     =   -5/ 2  (y   + 23/ 40) 

Now, compare this to the general form we provided:

(x -   h) ²  =  - 4p (y -  k)

From this we see that:  h = 3/4,  k =  -23/40  and 4p = 5/2  so: p = 5/8

Therefore, the following hold true and can be confirmed by sketching the graph (see problem 1):

a) The vertex is situated at V(3/4, - 23/40)

b) The axis of symmetry as always is x = h or x = 3/4 in this case.

c) The focus is p units below the vertex at F(3/4, - 6/5)


Problems:

1) Graph the parabola (x  - 3/4)²    =   -5/ 2  (y   + 23/ 40)   and defined by the properties given above.


2)The parabola shown below:


Has the general form  (y -   k) ²  =  - 4p (x  -  h)

Use this information to: a) find the actual equation of the curve, b) the coordinates of the focus, c) the equation for the directrix.

3) Find the focus and the directrix for the parabola: x²  =   8y

And graph it, clearly labeling the axis of symmetry.

Take the 2nd derivative before graphing and establish whether the parabola is oriented concave downward or upward.

4) Do the same for: y = x²  +   4x


5) Given  V(-3, 1) and F (0, 1) find the equation of the associated parabola and also for its directrix. Sketch the graph showing the focus, vertex and directrix.

6)  Find the tangent to the parabola y  =   x²     using the fact that the equation of the line tangent to a curve y = f(x) is given by:

y - y1 = f'(x1) (x - x1)

where (x1, y1) = P1 is the point of the curve for which the tangent is constructed and y1  =   (x1)²     .  Show the graph of the given parabola and the tangent line with the point P1 (x1, y1) identified.