Mathematics
Conics

1. The simple expression of a conic

The general expression of a conic is written as:

A x² + B x y + C y² + D x + E y + F = 0       (1)

The vector position r has the following coordinates in the frame xOy

r = x i + y j

A rotation of an angle θ of the frame xOy yields a new coordinates for the unit vectors i and j, as well as the vector position r. We have:

r = x' i' + y' j'
= x'(cos θ i + sin θ j) + y'(-sinθ i + cos θ) j
= (x'cos θ - y'sinθ)i + (x' sin θ + y' cos θ) j


Therefore:

x = x'cos θ - y'sinθ           (2)
y = x' sin θ + y' cos θ           (3)


The equation (1) becomes:

A (x'cos θ - y'sinθ)² + B (x'cos θ - y'sinθ)( x' sin θ + y' cos θ) + C (x' sin θ + y' cos θ)² + D (x'cos θ - y'sinθ) + E( x' sin θ + y' cos θ) + F = 0

or:

A (x'²cos² θ - 2 x'y' cos θ sinθ + y'²sin² θ) + B (x'² cos θ sin θ + x'y' cos² θ + - x'y'sin² θ - y'² sinθ cos θ)+ C (x'² sin² θ + y'² cos² θ)+ 2 x'y' sin θ cos θ)+ D x'cos θ - D y'sinθ + E x' sin θ + E y' cos θ + F = 0

or:

x'²[A cos²θ + B cos θ sin θ + C sin² θ] + x'y' [- 2 A cos θ sinθ + B cos² θ - B sin² θ + 2 C sin θ cos θ]+ y'²[A sin² θ) - B sinθ cos θ + C cos² θ] + x' (D cos θ + E sin θ) + y'( - D sinθ) + E cos θ) + F = 0

We can write then:

A' = A cos²θ + B cos θ sin θ + C sin² θ
B' = - 2 A cos θ sinθ + B cos² θ - B sin² θ + 2 C sin θ cos θ
C' = A sin² θ - B sinθ cos θ + C cos² θ
D' = D cos θ + E sin θ
E' = - D sinθ) + E cos θ
F' = F

Therefore:

A' x'² + B' x'y' + C'y'² + D' x' + E' y' + F' = 0           (4)

The factor of x'y' can be eliminated if B' = 0, so:

- 2 A cos θ sinθ + B cos² θ - B sin² θ + 2 C sin θ cos θ = 0

or:

B cos² θ - B sin² θ = 2 A cos θ sinθ - 2 C sin θ cos θ

B (cos 2θ) = 2( A - C) sin θ cos θ
B (cos 2θ) = ( A - C) sin 2θ

Thus:

tan 2θ = B/( A - C)

tan 2θ = B/( A - C)           (5)

A rotation of the frame xOy by an angle such as tan 2θ = B/( A - C), transforms the general form of a conic to the simple form (without the product xy)

The equation (4) becomes:

A' x'² + C'y'² + D' x' + E' y' + F' = 0           (6)

which is possible to write as X²/α² +/- Y² /β² = 1

2. The related forms of a conic

The equation of an ellipse have to be written as: x²/a² + y² /b² = 1 To have this, the product A'C' in the equation (6) must be positive. (A' and C' have the same sign)

The equation of an hyperbola have to be written as: x²/a² - y² /b² = 1 To have this, the product A'C' in the equation (6) must be negative. (A' and C' have different signs)

The equation of a parabola have to be written as: x² + α x + β y + γ = 0 or y² + α x + β y + γ = 0 To have this, the product A'C' in the equation (6) must be null. (A' or C' are equal to zero)

The are the three criteria to state of the form of the conic.

What is the expression of the product A'C' with respect to the first coefficients in the equation (1)?

A' = A cos²θ + B cos θ sin θ + C sin² θ
C' = A sin² θ - B sinθ cos θ + C cos² θ
A'C' = A² sin² θ) cos²θ - A B sinθ cos³θ + AC cos⁴ θ+ AB cos θ sin³ θ - B² sin²θ cos² θ + BC cos³ θsinθ + AC sin⁴ θ - BC sin³θ cos θ + C² sin² θ cos² θ

A'C' = (A² - B² + C²)sin² θ cos²θ (BC- A B )sinθ cos³θ + AC ( sin⁴ θ + cos⁴ θ) + (AB - BC) cos θ sin³ θ

we have:

sin² θ cos²θ = sin² 2θ /4

And:

sin⁴ θ + cos⁴ θ = 1 - 2 sin² θ cos²θ =
1 - sin² 2θ /2 = (1 + cos ² 2θ)/2

And:

(BC - A B) sinθ cos³θ + (AB - BC) cosθ sin³ θ
= (AB - BC) (sin² θ - cos²θ) sinθ cosθ
= - (AB - BC) (cos²θ - sin²θ)sinθ cosθ
= - (AB - BC) [cos 2θ sin 2θ ] /2

From the equation (5): tan 2θ = B/( A - C) , we have:

(BC - A B) sinθ cos³θ + (AB - BC) cosθ sin³ θ
= - B(A - C) [cos 2θ sin 2θ ] /2 = - B² [cos² 2θ ] /2

Then:

A'C' = (A² - B² + C²)sin² 2θ /4 + AC (1 + cos ² 2θ)/2 - B² [cos² 2θ ] /2

A'C' = (A² - B² + C²)sin² 2θ /4 + (AC/2 - B²/2 ) cos ² 2θ + AC/2

4 A'C' = (A² - B² + C²)sin² 2θ + 2(AC - B² ) cos ² 2θ + 2AC
= (A² - B² + C²)sin² 2θ + 2(AC - B²) (1 - sin ² 2θ) + 2AC
= (A² - B² + C²)sin² 2θ + 2(AC - B²) - 2(AC - B²) sin ² 2θ + 2AC
= (A² - B² + C² - 2AC + 2B² ) sin ² 2θ + 4AC - 2B²
= ( (A - C)² + B² ) sin ² 2θ + 4AC - 2B²

We have:

(A - C)² + B² = B²/tan² 2θ + B² = B²(1 + 1/tan² 2θ) = B²/sin² 2θ

Then:

4 A'C'= B²/sin² 2θ x sin ² 2θ + 4AC - 2B²
= B² + 4AC - 2B² = 4AC - B²

Therefore: - 4A'C' = B² - 4AC

- 4A'C' = B² - 4AC

Since B' = 0, we can write:

B'² - 4A'C' = B² - 4AC. This distriminant is invariant.

The criteria for the forms of conics from the product A'C', can be then directly used from the first form of the equation (1), that is B² - 4AC, before even proceeding to rotate the frame xOy.

Therefore, for the equation:

A x² + B x y + C y² + D x + E y + F = 0

• If B² - 4AC < 0, the equation represents an ellipse,
• If A = C and B = 0, the equation represents a circle,
• If B2 - 4AC = 0, the equation represents a parabola,
• If B2 - 4AC > 0, the equation represents a hyperbola.