Thursday, February 23, 2017

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