Solutions To Analytic Geometry Revisited (1)

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

Solutions:

a)  x  ²   +    y ²  - 2 y   =  3

We have: A = 1, D = 0,  E = -2  and F = -3

Then the center is at: 

(h, k) =    (-D/ 2A),   (-E/ 2A)  =   (0,  1)

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

So: 

r ²     = (0 ²   +    (-2) ²      - 4(-3_) /  4 (1) ²    

r ²     =   16/ 4   = 4  so radius  r  = 2

Sketch of the resulting circle:

b) 2x ²    +  2  y ²   + x  + y = 0

Here: A = 2,   C= 2,  D = 1,  E = 1,  F = 0

Center is at:

(h, k) =   (-D/ 2A),   (-E/ 2A)  =   (- 1/4,  -1/4)

Radius:

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

So: 

r ²     = (1 ) ²    +    (1) ²      - 4(0) /  4 (2) ²    

r ²     =   2    -  0   =   2  

So:  r =   Ö 2   (Sketch below)

c) x ²    +    y ²  + 2x  = 8

We have: A = 1,  D= 2, E = 0, F = -8

Center is at:

(h, k) =   (-D/ 2A),   (-E/ 2A)  =   (- 2/ 2,  0 ) = (-1, 0)

Radius:

r ²      = (2 ²     - 4(-8)) /  4 A ²    

So: 

r ²     =    36/ 4   =  9

So; r =   Ö  9   =    3

The circle is as shown below

2) For ellipse:    We have:   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:

 

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. 

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:

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