Friday, February 22, 2019

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:
No photo description available.

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)

No photo description available.

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

No photo description available.

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:
 No photo description available.

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:

No photo description available.


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