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Parabola


Enviado por   •  26 de Octubre de 2014  •  Tareas  •  523 Palabras (3 Páginas)  •  185 Visitas

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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:

Delta = |a b d; b c f; d f g|

(2)

J = |a b; b c|

(3)

I = a+c

(4)

K = |a d; d g|+|c f; f g|.

(5)

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.

curve Delta J Delta/I K

coincident lines 0 0 0

ellipse (imaginary) !=0 >0 >0

ellipse (real) !=0 >0 <0

hyperbola !=0 <0

intersecting lines (imaginary) 0 >0

intersecting lines (real) 0 <0

parabola !=0 0

parallel lines (imaginary) 0 0 >0

parallel lines (real) 0 0 <0

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

[x; y] = [costheta sintheta; -sintheta costheta][x^'; y^']

(6)

= [x^'costheta+y^'sintheta; -x^'sintheta+y^'costheta],

(7)

so

x = x^'costheta+y^'sintheta

(8)

y = -x^'sintheta+y^'costheta

(9)

xy = -x^('2)costhetasintheta+x^'y^'(cos^2theta-sin^2theta)+y^('2)costhetasintheta

(10)

x^2 = x^('2)cos^2theta+2x^'y^'costhetasintheta+y^('2)sin^2theta

(11)

y^2 = x^('2)sin^2theta-2x^'y^'sinthetacostheta+y^('2)cos^2theta.

(12)

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

a^' = acos^2theta-2bcosthetasintheta+csin^2theta

(15)

b^' = b(cos^2theta-sin^2theta)+(a-c)sinthetacostheta

(16)

c^' = asin^2theta+2bsinthetacostheta+ccos^2theta

(17)

d^' = dcostheta-fsintheta

(18)

f^' = dsintheta+fcostheta

(19)

g^' = g.

(20)

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

b^' = b(cos^2theta-sin^2theta)-(c-a)sinthetacostheta

(21)

= bcos(2theta)-1/2(c-a)sin(2theta)=0.

(22)

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

sintheta = sqrt((1-L)/2)

(26)

costheta = sqrt((1+L)/2).

(27)

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

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