Quadratic Curve (original) (raw)

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The general bivariate quadratic curve can be written

ax^2+2bxy+cy^2+2dx+2fy+g=0. (1)

Define the following quantities:

Then the quadratics are classified into the types summarized in the following table (Beyer 1987). The real (nondegenerate) quadratics (the ellipse and its special case the circle, hyperbola, and parabola) correspond to the curves which can be created by the intersection of a plane with a (two-nappes)cone, and are therefore known as conic sections.

It is always possible to eliminate the xy cross term by a suitable rotation of the axes. To see this, consider rotation by an arbitrary angle theta. The rotation matrix is

so

Plugging these into (◇) and grouping terms gives

x^('2)(acos^2theta+csin^2theta-2bcosthetasintheta)+x^'y^'[2acosthetasintheta-2csinthetacostheta+2b(cos^2theta-sin^2theta)]+y^('2)(asin^2theta+ccos^2theta+2bcosthetasintheta)+x^'(2dcostheta-2fsintheta)+y^'(2dsintheta+2fcostheta)+g=0. (13)

Comparing the coefficients with (◇) gives an equation of the form

a^'x^('2)+2b^'x^'y^'+c^'y^('2)+2d^'x^'+2f^'y^'+g^'=0, (14)

where the new coefficients are

The cross term 2b^'x^'y^' can therefore be made to vanish by setting

For b^' to be zero, it must be true that

cot(2theta)=(c-a)/(2b)=K. (23)

The other components are then given with the aid of the identity

cos[cot^(-1)(x)]=x/(sqrt(1+x^2)) (24)

by defining

L=K/(sqrt(1+K^2)), (25)

so

Rotating by an angle

theta=1/2cot^(-1)((c-a)/(2b)) (28)

therefore transforms (◇) into

a^'x^('2)+c^'y^('2)+2d^'x^'+2f^'y^'+g^'=0. (29)

Completing the square,

a^'(x^('2)+(2d^')/(a^')x)+c^'(y^('2)+(2f^')/(c^')y^')+g^'=0 (30)
a^'(x^'+(d^')/(a^'))^2+c^'(y^'+(f^')/(c^'))^2=-g^'+(d^('2))/(a^')+(f^('2))/(c^'). (31)

Defining x^('')=x^'+d^'/a^',y^('')=y^'+f^'/c^', and g^('')=-g^'+d^('2)/a^'+f^('2)/c^' gives

a^'x^(''2)+c^'y^(''2)=g^(''). (32)

If g^('')!=0, then divide both sides by g^(''). Defining a^('')=a^'/g^('') and c^('')=c^'/g^('') then gives

a^('')x^(''2)+c^('')y^(''2)=1. (33)

Therefore, in an appropriate coordinate system, the general conic section can be written (dropping the primes) as

{ax^2+cy^2=1 a,c,g!=0; ax^2+cy^2=0 a,c!=0, g=0. (34)

Consider an equation of the form ax^2+2bxy+cy^2=1 where b!=0. Re-express this using t_1 and t_2 in the form

ax^2+2bxy+cy^2=t_1x^('2)+t_2y^('2). (35)

Therefore, rotate the coordinate system

[x^'; y^']=[costheta sintheta; -sintheta costheta][x; y], (36)

so

ax^2+2bxy+cy^2=t_1x^('2)+t_2y^('2) =t_1(x^2cos^2theta+2xycosthetasintheta+y^2sin^2theta)+t_2(x^2sin^2theta-2xysinthetacostheta+y^2cos^2theta) =x^2(t_1cos^2theta+t_2sin^2theta)+2xycosthetasintheta(t_1-t_2)+y^2(t_1sin^2theta+t_2cos^2theta) (37)

and

Therefore,

From (39) and (45),

(a-c)/b=((t_1-t_2)cos(2theta))/(1/2(t_1-t_2)sin(2theta))=2cot(2theta), (46)

the same angle as before. But

so

a-c=(t_1-t_2)/(sqrt(1+((2b)/(a-c))^2)). (50)

Rewriting and copying (◇),

Adding (52) and (53) gives

Note that these roots can also be found from

(t-t_1)(t-t_2)=t^2-t(t_1+t_2)+t_1t_2=0 (56)

The original problem is therefore equivalent to looking for a solution to

[a b; b c][x; y]=t[x; y] (61)
[ax bx; by cy][x; y]=t[x^2; y^2], (62)

which gives the simultaneous equations

{ax^2+bxy=tx^2 ; bxy+cy^2=ty^2. (63)

Let X be any point (x,y) with old coordinates and (x^',y^') be its new coordinates. Then

ax^2+2bxy+cy^2=t_+x^('2)+t_-y^('2)=1 (64)

and

If t_+ and t_- are both >0, the curve is an ellipse. If t_+ and t_- are both <0, the curve is empty. If t_+ and t_- have opposite signs, the curve is a hyperbola. If either is 0, the curve is a parabola.

To find the general form of a quadratic curve in polar coordinates (as given, for example, in Moulton 1970), plug x=rcostheta and y=rsintheta into (◇) to obtain

ar^2cos^2theta+2br^2costhetasintheta+cr^2sin^2theta+2drcostheta+2frsintheta+g=0 (67)
(acos^2theta+2bcosthetasintheta+csin^2theta)+2/r(dcostheta+fsintheta)+g/(r^2)=0. (68)

Define u=1/r. For g!=0,we can divide through by 2g,

1/2u^2+1/g(dcostheta+fsintheta)u+1/(2g)(acos^2theta+2bcosthetasintheta+csin^2theta)=0. (69)

Applying the quadratic formula gives

u=-d/gcostheta-f/gsintheta+/-sqrt(R), (70)

where

Using the trigonometric identities

it follows that

Defining

then gives the equation

u=1/r=Asintheta+Bcostheta+/-sqrt(Csin(2theta)+Dcos(2theta)+E) (83)

(Moulton 1970). If g=0, then (◇) becomes instead

u=1/r=-(acos^2theta+2bcosthetasintheta+csin^2theta)/(2(dcostheta+fsintheta)). (84)

Therefore, the general form of a quadratic curve in polar coordinates is given by

u={Asintheta+Bcostheta for g!=0; +/-sqrt(Csin(2theta)+Dcos(2theta)+E) ; -(acos^2theta+2bcosthetasintheta+csin^2theta)/(2(dcostheta+fsintheta)) for g=0. (85)

See also

Algebraic Curve, Conic Section, Cubic Curve, Elliptic Curve, Quadratic, Quadratic Curve Discriminant

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References

Beyer, W. H. CRC Standard Mathematical Tables, 28th ed. Boca Raton, FL: CRC Press, pp. 200-201, 1987.Casey, J. "The General Equation of the Second Degree." Ch. 4 in A Treatise on the Analytical Geometry of the Point, Line, Circle, and Conic Sections, Containing an Account of Its Most Recent Extensions, with Numerous Examples, 2nd ed., rev. enl. Dublin: Hodges, Figgis, & Co., pp. 151-172, 1893.Moulton, F. R. "Law of Force in Binary Stars" and "Geometrical Interpretation of the Second Law." §58 and 59 in An Introduction to Celestial Mechanics, 2nd rev. ed. New York: Dover, pp. 86-89, 1970.

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Quadratic Curve

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Weisstein, Eric W. "Quadratic Curve." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/QuadraticCurve.html

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