Parabola

In mathematics, many of 2D shapes could be seen as the sections by intersection of a 3D object with a plane. For example, intersection of a sphere results in a circle. A square could be obtained by intersecting a cube. These intersections are quite straightforward. However, a parabola is such conic section that is not obvious to determine. A parabola is a conic section that is formed by intersecting a right circular cone with a plane at an angle that is parallel to one of the elements of the cone.

Definition as a locus of points

A parabola can also be defined as a set of points, in a plane, such that any point P in the set are equidistant between a fixed point F (focus) and a line l (directrix). The focus is a point located on the line of the axis of symmetry, while the directrix is a line perpendicular to the axis of symmetry. Sets for all the points on a parabola can be described by the equidistant property inside curly braces.

{\{P:|PF|=|Pl|\}.}

Drag the point P so that it moves along the curve of the parabola. It shows wherever P moves on the parabola, the distance between point F and point F is always equal to that to the line l.

The midpoint V of the perpendicular from the focus F onto the directrix l is called vertex, and the line FV is the axis of symmetry of the parabola.

Since the plane which intersects the cone is parallel to the generating line of the coen, a parabola is always unbounded U-shaped curve, which either opens upwards, downwards or in other different directions.

Parabola Equation

To simplify the process, let's start from the case when a parabola opens upwards, vertical parabola. The equation of the parabola Cartesian coordinate system can be expressed in the form of quadratic equation.

y=ax^2+bx+c

(1)

where a> 0 in our case.

Transform the equation (1) to get the vertex form of the quadratic equation.

y = (x+\dfrac{b}{2a})^2+\dfrac{4ac-b^2}{4a}

(2)

Then the coordinate for the vertex of the parabola is otained as below,

\Big( \dfrac{-b}{2a},\dfrac{4ac-b^2}{4a} \Big)

(3)

However, when it comes to study the properties of a parabola, the following equation is quite poular.

(x-h)^2=4p(y-k)

(4)

where,

h is x-coordinate of the vertex V

k is y-coordinate of the vertex V

p is focal length of the parabola, the length between the vertex V and the focus F. p\ne0. If p >0, the parabola opens upwards. If p < 0, the parabola opens downwards.

Comparing the equation (1) and (4) could determine the length of focal distance of a parabola given in the form of a quadratic equation.

Expand the equation (4) to its quadratic form,

y = \dfrac{1}{4p}x^2-\dfrac{h}{2p}x+\dfrac{h^2+4pk}{4p}

(5)

Compare the above equation with (1), we get the focal length p in terms of a.

p = \dfrac{1}{4a}

(6)

From the coordinates of vertex V (5), it's obvious h and k can be expressed in terms of a, b,c.

h = \dfrac{-b}{2a}

(7)

k = \dfrac{4ac-b^2}{4a}

(8)

Any chord of a parabola which passes through the focus is called the focal chord. The focal chord that is perpendicular to the axis of a parabola is called the latus rectum. The length of latus rectum of a parabola is equal to four times of its focal length.

In the figure, since the latus rectum CD is parallel to the axis x, the length of CD is the difference of x-coordinates of point C and point D.

Substitute the y-coordinate of the focus k+p into (5).

k+p = \dfrac{1}{4p}x^2-\dfrac{h}{2p}x+\dfrac{h^2+4pk}{4p}

Simplify the equation

\dfrac{1}{4p}x^2-\dfrac{h}{2p}x+\dfrac{h^2+4pk}{4p}-(k+p) = 0

\dfrac{1}{4p}x^2-\dfrac{h}{2p}x+\dfrac{h^2+4pk-4pk-4p^2}{4p}=0

x^2-2hx+h^2-4p^2 = 0

(9)

Solve the equation using the roots formula for quadratic equation. 2 solutions are obtained.

\begin{cases} x_1 = h-2p \\ x_2 = h+2p \end{cases}

Therefore,

x_2-x_1 = 4p

Horizontal Parabola - Rotating from Vertical One

For a parabola whose axis of symmetry is perpendicular to axis x, it is vertical. If its axis of symmetry is parallel to axis x, it is horizontal. Horizontal parabolas could be taken as vertical parabolas rotating 90\degree either clockwise or counterclockwise.

Rotating the parabola (4) by θ clockwise gives

(u-h)^2=4p(v-k)

(10)

where

\begin{aligned} \begin{pmatrix} u \\ v \end{pmatrix} &=\begin{pmatrix} \cosθ & -\sinθ \\ \sinθ & \cosθ \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \\ &= \begin{pmatrix} x\cos\theta -y\sin\theta \\ x\sin\theta + y\cos\theta \end{pmatrix} \end{aligned}

Substituting u and v into equation (10) gives general formula for a parabola after rotating an angle of θ.

(x\cos\theta -y\sin\theta-h)^2=4p(x\sin\theta + y\cos\theta-k)

(11)

If θ=90\degree, the equation is simplified as

(y+h)^2 = 4p( x-k)

(12)

After rotation about the Origin, the focus is transformed to the position F' with a new coordinate (k+p,-h). The sign of y-coordinate of the focus F' becomes negative since the point falls into quadrant Ⅳ.

Why is the conic section parabola

Let’s come to first question: why does intersecting a right circular cone with a plane produces a parabola when the plane is at an angle that is parallel to one of the elements of the cone.

First let’s have a refresh of a right circular cone. A right circular cone could be formed by revolving a right-angled triangle about any of its legs. For example, the right circular cone in the figure could be taken as revolving the right triangle \triangle SGO about its leg SO for 360\degree. The leg SO then forms the axis of the cone. Obviously, axis of the cone is perpendicular to the base of the cone, the surface of the circle which is formed by another leg OG of the right triangle. Therefore, any plane that containing the axis of a right circular cone is perpendicular to its base. In our case,

\text{plane}\, SGH \perp \text{plane} \,GPHM

(13)

Now it comes to our exciting moment. Suppose we could insert a sphere inside the cone such that the sphere is tangent to the surface of the right circular cone. The sphere is called Dandelin sphere, which was named after Belgian mathematician Germinal Pierre Dandelin, who's considered the first to discover the modern concept to prove conic sections of parabola, ellipse and hyperbola. For ellipse and hyperbola, two Dandelin spheres are needed for the proof. For parabola, one Dandelin sphere is needed.

It should be recognized that the surface of the intersection circle of the Dandelin sphere and the right circular cone is parallel to the base of the cone. In other words, the plane containing the intersecting circle is parallel to the base of the cone.

Let \beta denote the plane that contains the intersecting circle, then

\beta \parallel \text{plane} \,GPHM

(14)

Considering the conditions of (13) and (14), we know

\text{plane}\,\beta \perp \text{plane}\, SGH

(15)

Next, let's use a plane denoted as \alpha to intersect the cone at an angle that is parallel to the line SG and tangent to the sphere at point A. The section produced by the cone and plane \alpha is a parabola and the tangent point A is considered as the focus of the parabola. Intersecting line of plane \alpha and \beta l is considered the directrix of the parabola. To verify the statement, we are going to show the definition of a parabola is true for the conic section.

Since the plane \alpha is parallel to SG, we observed

\alpha \perp \text{plane}\, SGH

(16)

This statement could be separately proved. Since the intersection points of the plane with the base M and P are a pair of symmetrical points. \triangle AMP is an isosceles triangle.

AO\perp MP

On the other hand,

MP\perp GH

Since MP is perpendicular to two intersecting lines GH and AO in the \text{plane}\, SGH,

MP \perp \text{plane}\, SGH

while

MP\in \alpha

If any plane passing a line that is perpendicular to another plane, the two planes are perpendicular to each other. Therefore, our observation (16) is true.

Now considering statements (15) and (16), if two intersecting planes are perpendicular to a third plane, their intersection is also perpendicular to that plane. The intersection line of \alpha and \beta is l. Therefore,

l\perp \text{plane}\, SGH

(17)

l is perpendicular to any line on \text{plane}\, SGH, including the intersection with plane \alpha OJ.

Therefore,

l\perp OJ

(18)

Since the Quadrilateral DGOJ is a parallelogram, opposite sides are equal.

Therefore,

OJ = DG

(19)

It is noticed that DG is equal to the difference of generating line of the big right circular cone SG and that of the small one SD. Similarly, CP is the difference of generating line of the big right circular cone SP and that of the small one SC.

Therefore,

DG=CP

(20)

Now let's look at the Dandelin Sphere. Since point PC and PA are both tangent to the sphere, they have equal length.

PC = PA

(21)

Finally, let's plot a line d in plane \alpha that passes point P and is parallel to line OJ. Line OJ, OP, d and l form another parallelogram.

Therefore,

d = OJ

(22)

Considering the statement (17), d is perpendicular to l.

d\perp l

(23)

which shows d is the distance from the point on the parabola P to the line l. And it can be proved that d is equal to the distance from P to point A by iteration of equations from (22), (19), (20), (21).

d = PA

(24)

Since there's no constraint about the height of the cone, the point P on the section is arbitrary, which shows that for any point P in the intersection, it is equidistant between a fixed point A (focus) and a line l(directrix).

Conic section intersection why parabola

Collected in the board: Conic sections

Steven Zheng posted 1 year ago