While we have examined separately the circle, ellipse, parabola and hyperbola it should also be noted that all are second degree curves - what we call "conic sections" - which can be portrayed as special cases of a general analytic equation, viz.

Ax

^{2}+ Bxy + Cy

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

_{}

^{}

For example, the circle template equation with center at (h, k) e.g.

(x - h)

^{2}+ (y - k )^{2}= r^{2}
Can be extracted from the general analytic equation by taking:

A = C = 1 B = 0, D = -2h

E = - 2k, F = h

^{2}+ k^{2}- r^{2}
The parabola specific equation: x

Is obtained by using:

A = 1, E = - 4p, and B = C = D = F = 0

Note that the terms: Ax

are the second degree or quadratic terms with Bxy the "cross product" term

Most problems to do with this term involve eliminating using a rotation of Cartesian axes such that:

(x)

(y) =

(cos Θ. - sin Θ) (x')

(sin Θ. + cos Θ) .(y')

Performing the matrix operation for rotation:

x = x' cos Θ - y' sin Θ

y = x' sin Θ. + y' cos Θ

What if Θ = 45 degrees?

Then cos (45) = sin (45) = 1 /

Then: x = x' /

y = x' /

Now, if we apply the rotation of axes equation to the general analytic equation for 2nd degree curves what do we get? It can be shown we have:

A'x'

With the following relations to the original coefficients:

A' = A cos

B' = B (cos

C = A sin

D' = D cos Θ + E sin Θ

E' = - D sin Θ + E' cos Θ

F' = F

The technique for getting rid of the cross product terms is pretty straightforward, given an angle or rotation Θ can always be found such that the new cross product term is eliminated. How to do this? Simply set B' = 0 in the 2nd equation in the set above and solve for the angle Θ. Of course, it helps to have at hand a couple of trig identities:

cos

And:

2 sin Θ cos Θ = sin 2Θ

So that, for example:

B' = B cos 2Θ + (C - A) sin 2Θ

So B' will vanish if we choose:

cot 2Θ = (A - C)/ B

x

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') /

y = (x' + y' ) /

And

3(x')

Divide through by 6:

(x')

which is an ellipse with foci on the y' - axis. This is sketched below:

Problems:

1) Use the general analytic equation to write specific equations for the

2) The discriminant: B

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

b) Show that for the ellipse: B

c) Show that for the hyperbola: B

3) Determine the equation to which:

x

reduces when the axes are rotated to eliminate the cross product term. Sketch the resulting curve.

4) Use the discriminant to classify each of the following curves as a circle, parabola, ellipse, or parabola:

a) x

b) x

c) x

d) x

^{2}= 4pyIs obtained by using:

A = 1, E = - 4p, and B = C = D = F = 0

Note that the terms: Ax

^{2}, Bxy , Cy^{2}are the second degree or quadratic terms with Bxy the "cross product" term

Most problems to do with this term involve eliminating using a rotation of Cartesian axes such that:

(x)

(y) =

(cos Θ. - sin Θ) (x')

(sin Θ. + cos Θ) .(y')

Performing the matrix operation for rotation:

x = x' cos Θ - y' sin Θ

y = x' sin Θ. + y' cos Θ

What if Θ = 45 degrees?

Then cos (45) = sin (45) = 1 /

**Ö**2Then: x = x' /

**Ö**2 - y' /**Ö**2 = (x' - y') /**Ö**2y = x' /

**Ö**2 -+ y' /**Ö**2 = ( x' + y' ) /**Ö**2Now, if we apply the rotation of axes equation to the general analytic equation for 2nd degree curves what do we get? It can be shown we have:

A'x'

^{2}+ B' x' y' + C' y'^{2}+ D' x' + E'y' + F' = 0With the following relations to the original coefficients:

A' = A cos

^{2}Θ + B cos Θ sin Θ + C sin^{2 }ΘB' = B (cos

^{2}Θ - sin^{2 }Θ) + 2 (C - A) sin Θ cos ΘC = A sin

^{2 }Θ - B sin Θ cos Θ + C cos^{2}ΘD' = D cos Θ + E sin Θ

E' = - D sin Θ + E' cos Θ

F' = F

The technique for getting rid of the cross product terms is pretty straightforward, given an angle or rotation Θ can always be found such that the new cross product term is eliminated. How to do this? Simply set B' = 0 in the 2nd equation in the set above and solve for the angle Θ. Of course, it helps to have at hand a couple of trig identities:

cos

^{2}Θ**-**sin^{2}Θ = sin 2ΘAnd:

2 sin Θ cos Θ = sin 2Θ

So that, for example:

B' = B cos 2Θ + (C - A) sin 2Θ

So B' will vanish if we choose:

cot 2Θ = (A - C)/ B

*Example Problem*: Eliminate the cross product term in the equation:x

^{2}**+ xy + y**^{2}^{ }^{}^{ }= 3And 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**2y = (x' + y' ) /

**Ö**2And

_{}^{}3(x')

^{2}**+ ( y')**^{2}^{ }^{}^{ }= 6Divide through by 6:

(x')

^{2}**/**2**+ ( y')**^{2}^{ }^{}^{ }/ 6 = 1which is an ellipse with foci on the y' - axis. This is sketched below:

Problems:

1) Use the general analytic equation to write specific equations for the

*ellipse*and*hyperbola*. Sketch the resulting curves to confirm your answers.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 = 0b) Show that for the ellipse: B

^{2 }- 4AC < 0c) Show that for the hyperbola: B

^{2 }- 4AC > 03) Determine the equation to which:

x

^{2}**+ xy + y**^{2}^{ }^{}^{ }= 1reduces when the axes are rotated to eliminate the cross product term. Sketch the resulting curve.

4) Use the discriminant to classify each of the following curves as a circle, parabola, ellipse, or parabola:

a) x

^{2}**+ y**^{2}^{ }^{}^{ }+ xy + x - y^{ }^{}^{ }= 3b) x

^{2}**- y**^{2}^{ }^{}^{ }= 1c) x

^{2}**+ 3 y**^{2}^{ }^{}^{ }- 3 xy - 6 y^{ }^{}^{ }= 7d) x

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