Parabola (mathematics)

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For other uses, see Parabola (mathematics) (disambiguation).
Parabolic curve showing directrix (L) and focus (F). The distance from any point on the parabola to the focus (PnF) equals the perpendicular distance from the same point on the parabola to the directrix (PnQn).
Parabolic curve showing chord (L), focus (F), and vertex (V). L is an arbitrary chord of the parabola perpendicular to its axis of symmetry, which passes through V and F. (The ends of the chord are not shown here.) The lengths of all paths Qn - Pn - F are the same, equalling the distance between the chord L and the directrix. (See previous diagram above.) This is similar to saying that a parabola is an ellipse, but with one focal point at infinity. It also directly implies, by the wave nature of light, that parallel light arriving along the lines Qn - Pn will be reflected to converge at F. A linear wavefront along L is concentrated, after reflection, to the one point where all parts of it have travelled equal distances and are in phase, namely F. No consideration of angles is required.
A parabola obtained as the intersection of a cone with a plane parallel to a straight line on its surface.

In mathematics, a parabola (pron.: /pəˈræbələ/; plural parabolae or parabolas, from the Greek παραβολή) is a conic section, created from the intersection of a right circular conical surface and a plane parallel to a generating straight line of that surface. Another way to generate a parabola is to examine a point (the focus) and a line (the directrix). The locus of points in that plane that are equidistant from both the line and point is a parabola. In algebra, parabolas are frequently encountered as graphs of quadratic functions, such as y=x^2.

The line perpendicular to the directrix and passing through the focus (that is, the line that splits the parabola through the middle) is called the "axis of symmetry". The point on the axis of symmetry that intersects the parabola is called the "vertex", and it is the point where the curvature is greatest. The distance between the vertex and the focus, measured along the axis of symmetry, is the "focal length". Parabolas can open up, down, left, right, or in some other arbitrary direction. Any parabola can be repositioned and rescaled to fit exactly on any other parabola — that is, all parabolas are similar.

Parabolas have the property that, if they are made of material that reflects light, then light which enters a parabola travelling parallel to its axis of symmetry is reflected to its focus, regardless of where on the parabola the reflection occurs. Conversely, light that originates from a point source at the focus is reflected ("collimated") into a parallel beam, leaving the parabola parallel to the axis of symmetry. The same effects occur with sound and other forms of energy. This reflective property is the basis of many practical uses of parabolas.

The parabola has many important applications, from automobile headlight reflectors to the design of ballistic missiles. They are frequently used in physics, engineering, and many other areas.

Contents

[ History

Parabolic compass designed by Leonardo da Vinci
Parabolae are conic sections
Parabola construction using parallelogram method. Click magnify icon for explanation.

The earliest known work on conic sections was by Menaechmus in the fourth century BC. He discovered a way to solve the problem of doubling the cube using parabolae. (The solution, however, does not meet the requirements imposed by compass and straightedge construction.) The area enclosed by a parabola and a line segment, the so-called "parabola segment", was computed by Archimedes via the method of exhaustion in the third century BC, in his The Quadrature of the Parabola. The name "parabola" is due to Apollonius, who discovered many properties of conic sections. It means "application", referring to "application of areas" concept, that has a connection with this curve, as Apollonius had proved.[1] The focus–directrix property of the parabola and other conics is due to Pappus.

Galileo showed that the path of a projectile follows a parabola, a consequence of uniform acceleration due to gravity.

The idea that a parabolic reflector could produce an image was already well known before the invention of the reflecting telescope.[2] Designs were proposed in the early to mid seventeenth century by many mathematicians including René Descartes, Marin Mersenne,[3] and James Gregory.[4] When Isaac Newton built the first reflecting telescope in 1668 he skipped using a parabolic mirror because of the difficulty of fabrication, opting for a spherical mirror. Parabolic mirrors are used in most modern reflecting telescopes and in satellite dishes and radar receivers.[5]

[ Equation in Cartesian coordinates

Let the directrix be the line x = −p and let the focus be the point (p, 0). If (x, y) is a point on the parabola then, by Pappus' definition of a parabola, it is the same distance from the directrix as the focus; in other words:

x+p=\sqrt{(x-p)^2+y^2}.

Squaring both sides and simplifying produces

y^2 = 4px\

as the equation of the parabola. By interchanging the roles of x and y one obtains the corresponding equation of a parabola with a vertical axis as

x^2 = 4py.\

The equation can be generalized to allow the vertex to be at a point other than the origin by defining the vertex as the point (h, k). The equation of a parabola with a vertical axis then becomes

(x-h)^{2}=4p(y-k).\,

The last equation can be rewritten

y=ax^2+bx+c\,

so the graph of any function which is a polynomial of degree 2 in x is a parabola with a vertical axis.

More generally, a parabola is a curve in the Cartesian plane defined by an irreducible equation — one that does not factor as a product of two not necessarily distinct linear equations — of the general conic form

A x^{2} + B xy + C y^{2} + D x + E y + F = 0 \,

with the parabola restriction that

B^{2} = 4 AC,\,

where all of the coefficients are real and where A and C are not both zero. The equation is irreducible if and only if the determinant of the 3×3 matrix

\begin{bmatrix} A & B/2 & D/2 \\ B/2 & C & E/2 \\ D/2 & E/2 & F \end{bmatrix}.

is non-zero: that is, if (AC - B2/4)F + BED/4 - CD2/4 - AE2/4 ≠ 0. The reducible case, also called the degenerate case, gives a pair of parallel lines, possibly real, possibly imaginary, and possibly coinciding with each other.[6]

[ Conic section and quadratic form

Cone with cross-sections
Cone with cross-sections

The diagram represents a cone with its axis vertical.[7] The point A is its apex. A horizontal cross-section of the cone passes through the points B, E, C, and D. This cross-section is circular, but appears elliptical when viewed obliquely, as is shown in the diagram. An inclined cross-section of the cone, shown in pink, is inclined from the vertical by the same angle, θ, as the side of the cone. According to the definition of a parabola as a conic section, the boundary of this pink cross-section, EPD, is a parabola. The cone also has another horizontal cross-section, which passes through the vertex, P, of the parabola, and is also circular, with a radius which we will call r. Its centre is V, and PK is a diameter. The chord BC is a diameter of the lower circle, and passes through the point M, which is the midpoint of the chord ED. Let us call the lengths of the line segments EM and DM x, and the length of PM y.

Thus:

BM=2y\sin{\theta}.   (The triangle BPM is isosceles.)
CM=2r.   (PMCK is a parallelogram.)

Using the intersecting chords theorem on the chords BC and DE, we get:

EM \cdot DM=BM \cdot CM

Substituting:

x^2=4ry\sin{\theta}

Rearranging:

y=\frac{x^2}{4r\sin{\theta}}

For any given cone and parabola, r and θ are constants, but x and y are variables which depend on the arbitrary height at which the horizontal cross-section BECD is made. This last equation is a simple quadratic one which describes how x and y are related to each other, and therefore defines the shape of the parabolic curve. This shows that the definition of a parabola as a conic section implies its definition as the graph of a quadratic function. Both definitions produce curves of exactly the same shape.

[ Focal length

It is proved below that if a parabola has an equation of the form y = ax2, where a is a constant, then a=\frac{1}{4f}, where f is its focal length. Comparing this with the last equation above shows that the focal length of the above parabola is r sin θ.

[ Position of the focus

If a line is perpendicular to the plane of the parabola and passes through the centre, V, of the horizontal cross-section of the cone passing through P, then the point where this line intersects the plane of the parabola is the focus of the parabola, which is marked F on the diagram. Angle VPF is complementary to θ, and angle PVF is complementary to angle VPF, therefore angle PVF is θ. Since the length of PV is r, this construction correctly places the focus on the axis of symmetry of the parabola, at a distance r sin θ from its vertex.

[ Other geometric definitions

A parabola may also be characterized as a conic section with an eccentricity of 1. As a consequence of this, all parabolae are similar, meaning that while they can be different sizes, they are all the same shape. A parabola can also be obtained as the limit of a sequence of ellipses where one focus is kept fixed as the other is allowed to move arbitrarily far away in one direction. In this sense, a parabola may be considered an ellipse that has one focus at infinity. The parabola is an inverse transform of a cardioid.

A parabola has a single axis of reflective symmetry, which passes through its focus and is perpendicular to its directrix. The point of intersection of this axis and the parabola is called the vertex. A parabola spun about this axis in three dimensions traces out a shape known as a paraboloid of revolution.

The parabola is found in numerous situations in the physical world (see below).

[ Equations

[ Cartesian

In the following equations h and k are the coordinates of the vertex, (h,k), of the parabola and p is the distance from the vertex to the focus and the vertex to the directrix.

[ Vertical axis of symmetry

(x - h)^2 = 4p(y - k) \,
y =\frac{(x-h)^2}{4p}+k\,
y = ax^2 + bx + c \,

where

a = \frac{1}{4p}; \ \ b = \frac{-h}{2p}; \ \ c = \frac{h^2}{4p} + k; \ \
h = \frac{-b}{2a}; \ \ k = \frac{4ac - b^2}{4a}.

Parametric form:

x(t) = 2pt + h; \ \ y(t) = pt^2 + k \,

[ Horizontal axis of symmetry

(y - k)^2 = 4p(x - h) \,
x =\frac{(y - k)^2}{4p} + h;\ \,
x = ay^2 + by + c \,

where

a = \frac{1}{4p}; \ \ b = \frac{-k}{2p}; \ \ c = \frac{k^2}{4p} + h; \ \
h = \frac{4ac - b^2}{4a}; \ \ k = \frac{-b}{2a}.

Parametric form:

x(t) = pt^2 + h; \ \ y(t) = 2pt + k \,

[ General parabola

The general form for a parabola is

(\alpha x+\beta y)^2 + \gamma x + \delta y + \epsilon = 0 \,

This result is derived from the general conic equation given below:

Ax^2 +Bxy + Cy^2 + Dx + Ey + F = 0 \,

and the fact that, for a parabola,

B^2=4AC \,.

The equation for a general parabola with a focus point F(u, v), and a directrix in the form

ax+by+c=0 \,

is

\frac{\left(ax+by+c\right)^2}{{a}^{2}+{b}^{2}}=\left(x-u\right)^2+\left(y-v\right)^2 \,

[ Latus rectum, semilatus rectum, and polar coordinates

In polar coordinates, a parabola with the focus at the origin and the directrix parallel to the y-axis, is given by the equation

r (1 + \cos \theta) = l \,

where l is the semilatus rectum: the distance from the focus to the parabola itself, measured along a line perpendicular to the axis of symmetry. Note that this equals the perpendicular distance from the focus to the directrix, and is twice the focal length, which is the distance from the focus to the vertex of the parabola.

The latus rectum is the chord that passes through the focus and is perpendicular to the axis of symmetry. It has a length of 2l.

[ Gauss-mapped form

A Gauss-mapped form: (\tan^2\phi,2\tan\phi) has normal (\cos\phi,\sin\phi).

[ Proof of the reflective property

Reflective property of a parabola

The reflective property states that, if a parabola can reflect light, then light which enters it travelling parallel to the axis of symmetry is reflected to the focus. This is derived from the wave nature of light in the caption to a diagram near the top of this article. This derivation is valid, but may not be satisfying to readers who would prefer a mathematical approach. In the following proof, the fact that every point on the parabola is equidistant from the focus and from the directrix is taken as axiomatic.

Consider the parabola y=x^2. Since all parabolas are similar, this simple case represents all others. The right-hand side of the diagram shows part of this parabola.

Construction and definitions

The point E is an arbitrary point on the parabola, with coordinates (x,x^2). The focus is F, the vertex is A (the origin), and the line FA (the y-axis) is the axis of symmetry. The line EC is parallel to the axis of symmetry, and intersects the x-axis at D. The point C is located on the directrix (which is not shown, to minimize clutter). The point B is the midpoint of the line segment FC.

Deductions

Measured along the axis of symmetry, the vertex, A, is equidistant from the focus, F, and from the directrix. Correspondingly, since C is on the directrix, the y-coordinates of F and C are equal in absolute value and opposite in sign. B is the midpoint of FC, so its y-coordinate is zero, so it lies on the x-axis. Its x-coordinate is half that of E, D, and C, i.e. \frac{{x}}{{2}}. The slope of the line BE is the quotient of the lengths of ED and BD, which is \frac{x^2}{\left(\frac{x}{2}\right)}, which comes to 2x.

But 2x is also the slope (first derivative) of the parabola at E. Therefore the line BE is the tangent to the parabola at E.

The distances EF and EC are equal because E is on the parabola, F is the focus and C is on the directrix. Therefore, since B is the midpoint of FC, triangles FEB and CEB are congruent (three sides), which implies that the angles marked \alpha are equal. (The angle above E is vertically opposite angle BEC.) This means that a ray of light which enters the parabola and arrives at E travelling parallel to the axis of symmetry will be reflected by the line BE so it travels along the line EF, as shown in red in the diagram (assuming that the lines can somehow reflect light). Since BE is the tangent to the parabola at E, the same reflection will be done by an infinitessimal arc of the parabola at E. Therefore, light that enters the parabola and arrives at E travelling parallel to the axis of symmetry of the parabola is reflected by the parabola toward its focus.

The point E has no special characteristics. This conclusion about reflected light applies to all points on the parabola, as is shown on the left side of the diagram. This is the reflective property.

[ Tangent bisection property

The above proof, and the accompanying diagram, show that the tangent BE bisects the angle FEC. In other words, the tangent to the parabola at any point bisects the angle between the lines joining the point to the focus, and perpendicularly to the directrix.

[ Alternative proofs

Parabola and tangent

The above proofs of the reflective and tangent bisection properties use a line of calculus. For readers who are not comfortable with calculus, the following alternative is presented.

In the diagram to the right, F is the focus of the parabola, and T and U lie on its directrix. P is an arbitrary point on the parabola. PT is perpendicular to the directrix, and the line MP bisects angle FPT. Q is another point on the parabola, with QU perpendicular to the directrix. We know that FP=PT and FQ=QU. Clearly, QT>QU, so QT>FQ. All points on the bisector MP are equidistant from F and T, but Q is closer to F than to T. This means that Q is to the "left" of MP, i.e. on the same side of it as the focus. The same would be true if Q were located anywhere else on the parabola (except at the point P), so the entire parabola, except the point P, is on the focus side of MP. Therefore MP is the tangent to the parabola at P. Since it bisects the angle FPT, this proves the tangent bisection property.

The logic of the last paragraph can be applied to modify the above proof of the reflective property. It effectively proves the line BE to be the tangent to the parabola at E if the angles \alpha are equal. The reflective property follows as shown previously.

[ Two tangent properties

Let the line of symmetry intersect the parabola at point Q, and denote the focus as point F and its distance from point Q as f. Let the perpendicular to the line of symmetry, through the focus, intersect the parabola at a point T. Then (1) the distance from F to T is 2f, and (2) a tangent to the parabola at point T intersects the line of symmetry at a 45° angle.[8]:p.26

[ Orthoptic property

Perpendicular tangents intersect on the directrix
Main article: Isoptic

If two tangents to a parabola are perpendicular to each other, then they intersect on the directrix. Conversely, two tangents which intersect on the directrix are perpendicular.

Proof

Without loss of generality, consider the parabola y=x^2. Suppose that two tangents contact this parabola at the points (p,p^2) and (q,q^2). Their slopes are 2p and 2q, respectively. Thus the equation of the first tangent is of the form y=2px+C, where C is a constant. In order to make the line pass through (p,p^2), the value of C must be -p^2, so the equation of this tangent is y=2px-p^2. Likewise, the equation of the other tangent is y=2qx-q^2. At the intersection point of the two tangents, 2px-p^2=2qx-q^2. Thus 2x(p-q)=p^2-q^2. Factoring the difference of squares, cancelling, and dividing by 2 gives x=\frac{p+q}{2}. Substituting this into one of the equations of the tangents gives an expression for the y-coordinate of the intersection point: y=2p\left(\frac{p+q}{2}\right)-p^2. Simplifying this gives y=pq.

We now use the fact that these tangents are perpendicular. The product of the slopes of perpendicular lines is -1, assuming that both of the slopes are finite. The slopes of our tangents are 2p and 2q,, so (2p)(2q)=-1, so pq=-\frac{1}{4}. Thus the y-coordinate of the intersection point of the tangents is given by y=-\frac{1}{4}. This is also the equation of the directrix of this parabola, so the two perpendicular tangents intersect on the directrix.

[ Dimensions of parabolas with axes of symmetry parallel to the y-axis

These parabolas have equations of the form y=ax^2+bx+c. By interchanging x and y, the parabolas' axes of symmetry become parallel to the x-axis.

[ Coordinates of the vertex

The x-coordinate at the vertex is x=-\frac{b}{2a}, which is found by differentiating the original equation y=ax^2+bx+c, setting the resulting dy/dx=2ax+b equal to zero (a critical point), and solving for x. Substitute this x-coordinate into the original equation to yield:

y=a\left (-\frac{b}{2a}\right )^2 + b \left ( -\frac{b}{2a} \right ) + c.

Simplifying:

=\frac{ab^2}{4a^2} -\frac{b^2}{2a} + c

Put terms over a common denominator

=\frac{b^2}{4a} -\frac{2\cdot b^2}{2\cdot 2a} + c\cdot\frac{4a}{4a}
=\frac{-b^2+4ac}{4a}
=-\frac{b^2-4ac}{4a}=-\frac{D}{4a}

where D is the discriminant, (b^2-4ac).

Thus, the vertex is at point

\left (-\frac{b}{2a},-\frac{D}{4a}\right ).

[ Coordinates of the focus

Since the axis of symmetry of this parabola is parallel with the y-axis, the x-coordinates of the focus and the vertex are equal. The coordinates of the vertex are calculated in the preceding section. The x-coordinate of the focus is therefore also -\frac{b}{2a}.

To find the y-coordinate of the focus, consider the point, P, located on the parabola where the slope is 1, so the tangent to the parabola at P is inclined at 45 degrees to the axis of symmetry. Using the reflective property of a parabola, we know that light which is initially travelling parallel to the axis of symmetry is reflected at P toward the focus. The 45-degree inclination causes the light to be turned 90 degrees by the reflection, so it travels from P to the focus along a line that is perpendicular to the axis of symmetry and to the y-axis. This means that the y-coordinate of P must equal that of the focus.

By differentiating the equation of the parabola and setting the slope to 1, we find the x-coordinate of P:

y=ax^2+bx+c,
\frac{dy}{dx}=2ax+b=1
\therefore x=\frac{1-b}{2a}

Substituting this value of x in the equation of the parabola, we find the y-coordinate of P, and also of the focus:

y=a\left(\frac{1-b}{2a}\right)^2+b\left(\frac{1-b}{2a}\right)+c
=a\left(\frac{1-2b+b^2}{4a^2}\right)+\left(\frac{b-b^2}{2a}\right)+c
=\left(\frac{1-2b+b^2}{4a}\right)+\left(\frac{2b-2b^2}{4a}\right)+c
=\frac{1-b^2}{4a}+c=\frac{1-(b^2-4ac)}{4a}=\frac{1-D}{4a}

where D is the discriminant, (b^2-4ac), as is used in the "Coordinates of the vertex" section.

The focus is therefore the point:

\left(-\frac{b}{2a},\frac{1-D}{4a}\right)

[ Axis of symmetry, focal length, and directrix

The above coordinates of the focus of a parabola of the form:

y=ax^2+bx+c

can be compared with the coordinates of its vertex, which are derived in the section "Coordinates of the vertex", above, and are:

\left(\frac{-b}{2a},\frac{-D}{4a}\right)

where D=b^2-4ac.

The axis of symmetry is the line which passes through both the focus and the vertex. In this case, it is vertical, with equation:

x=-\frac{b}{2a}.

The focal length of the parabola is the difference between the y-coordinates of the focus and the vertex:

f=\left(\frac{1-D}{4a}\right)-\left(\frac{-D}{4a}\right)
=\frac{1}{4a}

It is sometimes useful to invert this equation and use it in the form: a=\frac{1}{4f}. See the section "Conic section and quadratic form", above.

Measured along the axis of symmetry, the vertex is the midpoint between the focus and the directrix. Therefore, the equation of the directrix is:

y=-\frac{D}{4a}-\frac{1}{4a}=-\frac{1+D}{4a}

[ Length of an arc of a parabola

If a point X is located on a parabola which has focal length f, and if p is the perpendicular distance from X to the axis of symmetry of the parabola, then the lengths of arcs of the parabola which terminate at X can be calculated from f and p as follows, assuming they are all expressed in thee same units.

h=\frac{p}{2}
q=\sqrt{f^2+h^2}CC-BY-SA.