## 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: 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)

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)

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: 