*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:

Ax

^{2}+ Bxy + Cy

^{2}+ Dx + Ey + F = 0

*: Let A = 9, B= 0, C = 4, D= 36, E = -8 and F = 4*

__For ellipse__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:

_{}

^{}

*: Let A = -1, B= 0, C = 4, D = E = F = 0*

__For hyperbola__Then we get:

4 y 2

^{ }

^{}^{ }- x

^{2}

**+ 4 x = 0**

_{}

^{}

Graphing yields:

2) The discriminant: B

^{2 }- 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: B

^{2 }- 4AC = 0

Solution: Take the sample parabola: y

^{2}

^{}= 4 x

The general analytic equation is:

Ax

^{2}+ Bxy + Cy

^{2}+ Dx + Ey + F = 0

Then for this case: B= 0 (no xy term), and A= 0, C = 1

Therefore: B

^{2 }- 4AC = - 4 AC = - 4(0)(1) = 0

b) Show that for the ellipse: B

^{2 }- 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: B

^{2 }- 4AC = - 4 AC = - 4(9)(4) = -324

So the condition is satisfied

c) Show that for the hyperbola: B

^{2 }- 4AC > 0

Solution: Using the hyperbola from Problem 1:

4 y

^{2}

^{ }

^{}^{ }- x

^{2}

**+ 4 x = 0**

Therefore: B= 0, A = -1, C = 4

Then: B

^{2 }- 4AC = - 4 (-1)(4) = 16

Which is greater than zero so the condition holds.

3) Determine the equation to which:

x

^{2}

**+ xy + y**

^{2}

^{ }

^{}^{ }= 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) x

^{2}

**+ y**

^{2}

^{ }

^{}^{ }+ xy + x - y

^{ }

^{}^{ }= 3

A = 1, B = 1, C = 1

Then: B

^{2 }- 4AC = (1)

^{2}- 4 (1)(1) = 1 - 4 = -3

Since B

^{2 }- 4AC < 0 the curve is an

*ellipse*

b) x

^{2}

**- y**

^{2}

^{ }

^{}^{ }= 1

A = 1, B = 0, C = -1

Then: B

^{2 }- 4AC = - 4 (1)(-1) = 4

Since B

^{2 }- 4AC > 0 the curve is a

*hyperbola*

c) x

^{2}

**+ 3 y**

^{2}

^{ }

^{}^{ }- 3 xy - 6 y

^{ }

^{}^{ }= 7

A = 1, B = -3, C = 3

Then: B

^{2 }- 4AC = (-3)

^{2}- 4(1) (3) = 9 - 12 = -3

Since B

^{2 }- 4AC < 0 the curve is an

*ellipse*

_{}

^{}

d) x

^{2}

**+ y**

^{2}

^{ }

^{}^{ }+ 3 x - 2 y

^{ }

^{}^{ }= 3

A = 1, B = 0, C = 1

Then: B

^{2 }- 4AC = - 4 (1)(1) = - 4

Since B

^{2 }- 4AC < 0 the curve is an

*ellipse*

_{}

^{}

## No comments:

Post a Comment