Conic section

In mathematics, a conic section (or just conic) is a curved locus of points, formed by intersecting a cone with a plane. The conic sections were named and studied as long ago as 200 BC, when Apollonius of Perga undertook a systematic study of their properties.

Derivation

Let there be a cone whose axis is the z-axis. Let its vertex be the origin. The equation for the cone is

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: x^2 + y^2 - a^2 z^2 = 0 qquad qquad (1)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

where

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: a = an heta > 0 ;

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

and heta is the angle which the generators of the cone make with respect to the axis. Notice that this cone is actually a pair of cones: one cone standing upside down on the vertex of the other cone—or, as mathematicians say, this cone consists of two "nappes."

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Let there be a plane with a slope running along the x direction but which is level along the y direction. Its equation is

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: z = mx + b qquad qquad (2)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

where

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: m = an phi > 0 ;

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

and phi is the angle of the plane with respect to the x-y plane.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We are interested in finding the intersection of the cone and the plane, which means that equations (1) and (2) shall be combined. Both equations can be solved for z and then equate the two values of z. Solving equation (1) for z yields

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: z = sqrt{x^2 + y^2 over a^2}

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

therefore

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: sqrt{x^2 + y^2 over a^2} = m x + b.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Square both sides and expand the squared binomial on the right side,

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: {x^2 + y^2 over a^2} = m^2 x^2 + 2 m b x + b^2. ;

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Grouping by variables yields

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: x^2 left( {1 over a^2} - m^2 ight) + {y^2 over a^2} - 2 m b x - b^2 = 0. qquad qquad (3)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Note that this is the equation of the projection of the conic section on the xy-plane, hence contracted in the x-direction compared with the shape of the conic section itself.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Derivation of the parabola

The parabola is obtained when the slope of the plane is equal to the slope of the generators of the cone. When these two slopes are equal, then the angles heta and phi become complementary. This implies that

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: an heta = cot phi ;

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

therefore

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: m = {1 over a}. qquad qquad (4)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Substituting equation (4) into equation (3) makes the first term in equation (3) vanish, and the remaining equation is

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: {y^2 over a^2} - {2 over a} b x - b^2 = 0.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Multiply both sides by a2,

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: y^2 - 2 a b x - a^2 b^2 = 0 ;

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

then solve for x,

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: x = {1 over 2 a b} y^2 - {a b over 2}. qquad qquad (5)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Equation (5) describes a parabola whose axis is parallel to the x-axis. Other versions of equation (5) can be obtained by rotating the plane around the z-axis.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Derivation of the ellipse

An ellipse happens when the angles heta and phi, when added together, do not measure up to a right angle:

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: heta + phi < {pi over 2} qquad qquad mbox{(ellipse)}

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

which implies that the tangent of the sum of these two angles is positive.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: an ( heta + phi) > 0. ;

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

But a trigonometric identity states that

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: an ( heta + phi) = { an heta + an phi over 1 - an heta an phi}

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

therefore

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: an ( heta + phi) = {m + a over 1 - m a} > 0 qquad qquad (6)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

but m + a is positive, since the summands are given to be positive, so inequality (6) is positive if the denominator is also positive:

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: 1 - m a > 0. qquad qquad (7)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

From inequality (7) we can deduce

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: m a < 1, ;

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: m^2 a^2 < 1, ;

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: 1 - m^2 a^2 > 0, ;

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: {1 over m^2 a^2} > 1,

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: {1 over m^2 a^2} - 1 > 0,

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: {1 over a^2} - m^2 > 0 qquad qquad mbox{(ellipse)}.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Let us start out again from equation (3),

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: x^2 left( {1 over a^2} - m^2 ight) + {y^2 over a^2} - 2 m b x - b^2 = 0, qquad qquad (3)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

but this time the coefficient of the x2 term does not vanish but is instead positive. Solve for y,

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: y = a sqrt{ b^2 + 2 m b x - x^2 left( {1 over a^2} - m^2 ight)}. qquad qquad (8)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

This would clearly describe an ellipse were it not for the second term under the radical, the 2 m b x: it would be the equation of a circle which has been stretched proportionally along the directions of the x-axis and the y-axis. Equation (8) is an ellipse but it is not obvious, so it will be rearranged further until it is obvious. Complete the square under the radical,

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: y = a sqrt{ b^2 - left^2 + left( {b^2 over {1 over a^2 m^2} - 1} ight)}.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Group together the b2 terms,

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: y = a sqrt{ b^2 left( 1 + {1 over {1 over a^2 m^2} - 1} ight) - left^2 }.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Divide by a then square both sides,

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: {y^2 over a^2} + left( x sqrt{{1 over a^2} - m^2} - {b over sqrt{{1 over a^2 m^2} - 1}} ight)^2 = b^2 left( 1 + {1 over {1 over a^2 m^2} - 1} ight).

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The x has a coefficient. It is desired to pull this coefficient out by factoring it out of the second term which is a square,

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: {y^2 over a^2} + left( {1 over a^2} - m^2 ight) left( x - {b over sqrt{ left( {1 over a^2 m^2} - 1 ight) left( {1 over a^2} - m^2 ight)}} ight)^2 = b^2 left( 1 + { 1 over {1 over a^2 m^2} - 1} ight).

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Further rearrangements of constants finally leads to

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: {y^2 over 1 - a^2 m^2} + left( x - {m b over {1 over a^2} - m^2} ight)^2 = { a^2 b^2 over (1 - a^2 m^2)^2}.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The coefficient of the y term is positive (for an ellipse). Renaming of coefficients and constants leads to

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: {y^2 over A} + (x - C)^2 = R^2 qquad qquad (9)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

which is clearly the equation of an ellipse. That is, equation (9) describes a circle of radius R and center (C,0) which is then stretched vertically by a factor of sqrt{A} . The second term on the left side (the x term) has no coefficient but is a square, so that it must be positive. The radius is a product of squares, so it must also be positive. The first term on the left side(the y term) has a coefficient which is positive, so the equation describes an ellipse.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Derivation of the hyperbola

The hyperbola happens when the angles heta and phi add up to an obtuse angle, which is greater than a right angle. The tangent of an obtuse angle is negative. All the inequalities which were valid for the ellipse become reversed. Therefore

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

: 1 - a^2 m^2 < 0 qquad qquad mbox{(hyperbola)}.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Otherwise the equation for the hyperbola is the same as equation (9) for the ellipse, except that the coefficient A of the y term is negative. The sign change is enough to convert an ellipse into a hyperbola. This is because the equation of a real ellipse contains an imaginary hyperbola, and the equation of a real hyperbola contains an imaginary ellipse (see imaginary number). The sign change of coefficient A causes real and imaginary values of the function y=f(x) equivalent to equation (9) to swap.

Related Topics:
Hyperbola - Imaginary number

~ ~ ~ ~ ~ ~ ~ ~ ~ ~