Math - Rotational Conics

ROTATIONAL CONICS

Given that:

Therefore, x = r cosθ = r cos(θ-α)

and y = r sinθ and

= r sin(θ-α)

Using the General Addition Formulas1, 2, we have: = x cosα + y sinα = y cosα – x sinα

and

x=

cosα -

sinα

Rotation Transformation

y=

sinα +

cosα

Equations

To get the angle of rotation of the plane in which the xy coefficient disappears, we equate the coefficients of the transformed general equation to zero. We get3: cot2α = A-C B

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To know the graph of the tranformed equation beforehand, we make use of the discriminant. Take note that4: 2

- 4Ā = B2 - 4AC

And so, If Ā > 0, then it is an ellipse, with B2 – 4AC < 0 Ā = 0, either Ā=0 or =0, then it is a parabola, with B2 – 4AC = 0 Ā < 0, then it is a hyperbola, with B2 – 4AC > 0

PROVING 1

2

3

= r cos(θ-α)

= r sin(θ-α)

= r (cosθcosα + sinθsinα)

= r (sinθcosα - cosθsinα)

= (r cosθ)(cosα) + (sinθ)(r sinα)

= (r sinθ)(cosα) – (r cosθ)(sinα)

= x cosα + y sinα

= y cosα – x sinα

x = r cosθ

y = r sinθ

= r ( (cosα)(cos(θ-α)) - (sinα)(sin(θ-α)) )

= r ( (sinα)(cos(θ-α)) + (cosα)(sin(θ-α)) )

= (r (cos(θ-α)))(cosα) – (r (sin(θ-α)))(sinα)

= (r (cos(θ-α)))(sinα) + (r (sin(θ-α)))(cosα)

=

=

cosα -

sinα

Ax2 = Acos2α

2

– 2Asinαcosα

Bxy=Bsinαcosα( Cy2 = Csin2α

2

2

-

2

)+

+ 2Csinαcosα

+ Asin2α

2

B(cos2α-sin2α) + Ccos2α

2

sinα +

cosα

0 = B(cos2α - sin2α) + (C-A)(2sinαcosα) = Bcos2α + (C-A)(sin2α) Bcos2α = (A-C)(sin2α) cos2α/sin2α = (A-C)/B cot2α = (A-C)/B

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4

From the transformation of general equation (above), Ā = Acos2α + Bsinαcosα + Csin2 α

= Asin2α - Bsinαcosα + Ccos2α

= B(cos2α - sin2α) + (C-A)(2sinαcosα)

Then, B2-4AC

2

- 4Ā

(B(cos2α - sin2α) + (C-A)(2sinαcosα))2 – 4(Acos2α + Bsinαcosα + Csin2α) (Asin2α - Bsinαcosα + Ccos2α) B2cos4α -2B2cos2αsin2α + B2sin4α +4(BC-AB)(sinαcos3α) +4(AB-BC)(sin3αcosα) + 4C2sin2αcos2α -8ACsin2αcos2α +4A2sin2αcos2α -4A2sin2 αcos2α +24B2sin2αcos2α -4C2sin2αcos2α -4AC(sin4α+cos4α) -4(BC-AB)(sinαcos3α) -4(AB-BC)(sin3αcosα) B2cos4α + B2sin4α -8ACsin2αcos2α +2B2sin2 αcos2α -4AC(sin4 α+cos4α) B2(cos4α+sin4α+2sin2αcos2α) – 4AC(2sin2αcos2α+ cos4α+sin4α) (B2-4AC)(cos2α+sin2α)2 (B2-4AC)(1)2 ≡ B2-4AC

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