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 the cross product 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 the cross product 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
## 2 comments:

Your picture of a parabola is an ellipse. It needs to be parallel to the side.

No, the graphic is quite correct. You're looking at it in the wrong perspective. The parabola is shown in accurate form as a cut in the plane given. It does not need to be parallel to the side. Indeed, similar diagrams as that shown will be find in any good text on analytic geometry.

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