Parabola

Parabolic compass designed by Leonardo da Vinci

The earliest known work on conic sections was by Menaechmus in the 4th century BC. He discovered a way to solve the problem of doubling the cube using parabolas. (The solution, however, does not meet the requirements of compass-and-straightedge construction.) The area enclosed by a parabola and a line segment, the so-called "parabola segment", was computed by Archimedes by the method of exhaustion in the 3rd 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 conic sections 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-17th 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]

A parabola can be defined geometrically as a set of points (locus of points) in the Euclidean plane:

• A parabola is a set of points, such that for any point ${\displaystyle P}$ of the set the distance ${\displaystyle |PF|}$ to a fixed point ${\displaystyle F}$, the focus, is equal to the distance ${\displaystyle |Pl|}$ to a fixed line ${\displaystyle l}$, the directrix:

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

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

Parabolas ${\displaystyle y=ax^{2}}$

The previous section shows that any parabola with the origin as vertex and the y axis as axis of symmetry can be considered as the graph of a function

${\displaystyle f(x)=ax^{2}{\text{ with }}a\neq 0.}$

For ${\displaystyle a>0}$ the parabolas are opening to the top, and for ${\displaystyle a<0}$ are opening to the bottom (see picture). From the section above one obtains:

• The focus is ${\displaystyle \left(0,{\frac {1}{4a}}\right)}$,
• the focal length ${\displaystyle {\frac {1}{4a}}}$, the semi-latus rectum is ${\displaystyle p={\frac {1}{2a}}}$,
• the vertex is ${\displaystyle (0,0)}$,
• the directrix has the equation ${\displaystyle y=-{\frac {1}{4a}}}$,
• the tangent at point ${\displaystyle (x_{0},ax_{0}^{2})}$ has the equation ${\displaystyle y=2ax_{0}x-ax_{0}^{2}}$.

For ${\displaystyle a=1}$ the parabola is the unit parabola with equation ${\displaystyle y=x^{2}}$. Its focus is ${\displaystyle \left(0,{\tfrac {1}{4}}\right)}$, the semi-latus rectum ${\displaystyle p={\tfrac {1}{2}}}$, and the directrix has the equation ${\displaystyle y=-{\tfrac {1}{4}}}$.

The general function of degree 2 is

${\displaystyle f(x)=ax^{2}+bx+c{\text{ with }}a,b,c\in \mathbb {R} ,\ a\neq 0}$.

Completing the square yields

${\displaystyle f(x)=a\left(x+{\frac {b}{2a}}\right)^{2}+{\frac {4ac-b^{2}}{4a}},}$

which is the equation of a parabola with

• the axis ${\displaystyle x=-{\frac {b}{2a}}}$ (parallel to the y axis),
• the focal length ${\displaystyle {\frac {1}{4a}}}$, the semi-latus rectum ${\displaystyle p={\frac {1}{2a}}}$,
• the vertex ${\displaystyle V=\left(-{\frac {b}{2a}},{\frac {4ac-b^{2}}{4a}}\right)}$,
• the focus ${\displaystyle F=\left(-{\frac {b}{2a}},{\frac {4ac-b^{2}+1}{4a}}\right)}$,
• the directrix ${\displaystyle y={\frac {4ac-b^{2}-1}{4a}}}$,
• the point of the parabola intersecting the y axis has coordinates ${\displaystyle (0,c)}$,
• the tangent at a point on the y axis has the equation ${\displaystyle y=bx+c}$.

When the parabola ${\displaystyle \color {blue}{y=2x^{2}}}$ is uniformly scaled by factor 2, the result is the parabola ${\displaystyle \color {red}{y=x^{2}}}$

Two objects in the Euclidean plane are similar if one can be transformed to the other by a similarity, that is, an arbitrary composition of rigid motions (translations and rotations) and uniform scalings.

A parabola ${\displaystyle {\mathcal {P}}}$ with vertex ${\displaystyle V=(v_{1},v_{2})}$ can be transformed by the translation ${\displaystyle (x,y)\to (x-v_{1},y-v_{2})}$ to one with the origin as vertex. A suitable rotation around the origin can then transform the parabola to one that has the y axis as axis of symmetry. Hence the parabola ${\displaystyle {\mathcal {P}}}$ can be transformed by a rigid motion to a parabola with an equation ${\displaystyle y=ax^{2},\ a\neq 0}$. Such a parabola can then be transformed by the uniform scaling ${\displaystyle (x,y)\to (ax,ay)}$ into the unit parabola with equation ${\displaystyle y=x^{2}}$. Thus, any parabola can be mapped to the unit parabola by a similarity.[6]

A synthetic approach, using similar triangles, can also be used to establish this result.[7]

The general result is that two conic sections (necessarily of the same type) are similar if and only if they have the same eccentricity.[6] Therefore, only circles (all having eccentricity 0) share this property with parabolas (all having eccentricity 1), while general ellipses and hyperbolas do not.

There are other simple affine transformations that map the parabola ${\displaystyle y=ax^{2}}$ onto the unit parabola, such as ${\displaystyle (x,y)\to \left(x,{\tfrac {y}{a}}\right)}$. But this mapping is not a similarity, and only shows that all parabolas are affinely equivalent (see § As the affine image of the unit parabola).

Pencil of conics with a common vertex

The pencil of conic sections with the x axis as axis of symmetry, one vertex at the origin (0, 0) and the same semi-latus rectum ${\displaystyle p}$ can be represented by the equation

${\displaystyle y^{2}=2px+(e^{2}-1)x^{2},\quad e\geq 0,}$

with ${\displaystyle e}$ the eccentricity.

• For ${\displaystyle e=0}$ the conic is a circle (osculating circle of the pencil),
• for ${\displaystyle 0<e<1}$ an ellipse,
• for ${\displaystyle e=1}$ the parabola with equation ${\displaystyle y^{2}=2px,}$
• for ${\displaystyle e>1}$ a hyperbola (see picture).

Pencil of conics with a common focus

If p > 0, the parabola with equation ${\displaystyle y^{2}=2px}$ (opening to the right) has the polar representation

${\displaystyle r=2p{\frac {\cos \varphi }{\sin ^{2}\varphi }},\quad \varphi \in \left[-{\tfrac {\pi }{2}},{\tfrac {\pi }{2}}\right]\setminus \{0\}}$

(${\displaystyle r^{2}=x^{2}+y^{2},\ x=r\cos \varphi }$).

Its vertex is ${\displaystyle V=(0,0)}$, and its focus is ${\displaystyle F=\left({\tfrac {p}{2}},0\right)}$.

If one shifts the origin into the focus, that is, ${\displaystyle F=(0,0)}$, one obtains the equation

${\displaystyle r={\frac {p}{1-\cos \varphi }},\quad \varphi \neq 2\pi k.}$

Remark 1: Inverting this polar form shows that a parabola is the inverse of a cardioid.

Remark 2: The second polar form is a special case of a pencil of conics with focus ${\displaystyle F=(0,0)}$ (see picture):

${\displaystyle r={\frac {p}{1-e\cos \varphi }}}$ (${\displaystyle e}$ is the eccentricity).

Reflective property of a parabola

The reflective property states that if a parabola can reflect light, then light that enters it travelling parallel to the axis of symmetry is reflected toward the focus. This is derived from geometrical optics, based on the assumption that light travels in rays.

Consider the parabola y = x². Since all parabolas are similar, this simple case represents all others.

Intersection of a tangent and perpendicular from focus

Perpendicular from focus to tangent

Since triangles △FBE and △CBE are congruent, FB is perpendicular to the tangent BE. Since B is on the x axis, which is the tangent to the parabola at its vertex, it follows that the point of intersection between any tangent to a parabola and the perpendicular from the focus to that tangent lies on the line that is tangential to the parabola at its vertex. See animated diagram[8] and pedal curve.

Alternative proofs

Parabola and tangent

The above proofs of the reflective and tangent bisection properties use a line of calculus. Here a geometric proof is presented.

In this diagram, 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, that is, 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 α are equal. The reflective property follows as shown previously.

Parabola: pin string construction

The definition of a parabola by its focus and directrix can be used for drawing it with help of pins and strings:[9]

1. Choose the focus ${\displaystyle F}$ and the directrix ${\displaystyle l}$ of the parabola.
2. Take a triangle of a set square and prepare a string with length ${\displaystyle |AB|}$ (see diagram).
3. Pin one end of the string at point ${\displaystyle A}$ of the triangle and the other one to the focus ${\displaystyle F}$.
4. Position the triangle such that the second edge of the right angle is free to slide along the directrix.
5. Take a pen and hold the string tight to the triangle.
6. While moving the triangle along the directrix, the pen draws an arc of a parabola, because of ${\displaystyle |PF|=|PB|}$ (see definition of a parabola).

A parabola can be considered as the affine part of a non-degenerated projective conic with a point ${\displaystyle Y_{\infty }}$ on the line of infinity ${\displaystyle g_{\infty }}$, which is the tangent at ${\displaystyle Y_{\infty }}$. The 5-, 4- and 3- point degenerations of Pascal's theorem are properties of a conic dealing with at least one tangent. If one considers this tangent as the line at infinity and its point of contact as the point at infinity of the y axis, one obtains three statements for a parabola.

The following properties of a parabola deal only with terms connect, intersect, parallel, which are invariants of similarities. So, it is sufficient to prove any property for the unit parabola with equation ${\displaystyle y=x^{2}}$.

4-points property

4-points property of a parabola

Any parabola can be described in a suitable coordinate system by an equation ${\displaystyle y=ax^{2}}$.

• Let ${\displaystyle P_{1}=(x_{1},y_{1}),\ P_{2}=(x_{2},y_{2}),\ P_{3}=(x_{3},y_{3}),\ P_{4}=(x_{4},y_{4})}$ be four points of the parabola ${\displaystyle y=ax^{2}}$, and ${\displaystyle Q_{2}}$ the intersection of the secant line ${\displaystyle P_{1}P_{4}}$ with the line ${\displaystyle x=x_{2},}$ and let ${\displaystyle Q_{1}}$ be the intersection of the secant line ${\displaystyle P_{2}P_{3}}$ with the line ${\displaystyle x=x_{1}}$ (see picture). Then the secant line ${\displaystyle P_{3}P_{4}}$ is parallel to line ${\displaystyle Q_{1}Q_{2}}$.

(The lines ${\displaystyle x=x_{1}}$ and ${\displaystyle x=x_{2}}$ are parallel to the axis of the parabola.)

Proof: straightforward calculation for the unit parabola ${\displaystyle y=x^{2}}$.

Application: The 4-points property of a parabola can be used for the construction of point ${\displaystyle P_{4}}$, while ${\displaystyle P_{1},P_{2},P_{3}}$ and ${\displaystyle Q_{2}}$ are given.

Remark: the 4-points property of a parabola is an affine version of the 5-point degeneration of Pascal's theorem.

3-points–1-tangent property

3-points–1-tangent property

Let ${\displaystyle P_{0}=(x_{0},y_{0}),P_{1}=(x_{1},y_{1}),P_{2}=(x_{2},y_{2})}$ be three points of the parabola with equation ${\displaystyle y=ax^{2}}$ and ${\displaystyle Q_{2}}$ the intersection of the secant line ${\displaystyle P_{0}P_{1}}$ with the line ${\displaystyle x=x_{2}}$ and ${\displaystyle Q_{1}}$ the intersection of the secant line ${\displaystyle P_{0}P_{2}}$ with the line ${\displaystyle x=x_{1}}$ (see picture). Then the tangent at point ${\displaystyle P_{0}}$ is parallel to the line ${\displaystyle Q_{1}Q_{2}}$. (The lines ${\displaystyle x=x_{1}}$ and ${\displaystyle x=x_{2}}$ are parallel to the axis of the parabola.)

Proof: can be performed for the unit parabola ${\displaystyle y=x^{2}}$. A short calculation shows: line ${\displaystyle Q_{1}Q_{2}}$ has slope ${\displaystyle 2x_{0}}$ which is the slope of the tangent at point ${\displaystyle P_{0}}$.

Application: The 3-points-1-tangent-property of a parabola can be used for the construction of the tangent at point ${\displaystyle P_{0}}$, while ${\displaystyle P_{1},P_{2},P_{0}}$ are given.

Remark: The 3-points-1-tangent-property of a parabola is an affine version of the 4-point-degeneration of Pascal's theorem.

2-points–2-tangents property

2-points–2-tangents property

Let ${\displaystyle P_{1}=(x_{1},y_{1}),\ P_{2}=(x_{2},y_{2})}$ be two points of the parabola with equation ${\displaystyle y=ax^{2}}$, and ${\displaystyle Q_{2}}$ the intersection of the tangent at point ${\displaystyle P_{1}}$ with the line ${\displaystyle x=x_{2}}$, and ${\displaystyle Q_{1}}$ the intersection of the tangent at point ${\displaystyle P_{2}}$ with the line ${\displaystyle x=x_{1}}$ (see picture). Then the secant ${\displaystyle P_{1}P_{2}}$ is parallel to the line ${\displaystyle Q_{1}Q_{2}}$. (The lines ${\displaystyle x=x_{1}}$ and ${\displaystyle x=x_{2}}$ are parallel to the axis of the parabola.)

Proof: straight forward calculation for the unit parabola ${\displaystyle y=x^{2}}$.

Application: The 2-points–2-tangents property can be used for the construction of the tangent of a parabola at point ${\displaystyle P_{2}}$, if ${\displaystyle P_{1},P_{2}}$ and the tangent at ${\displaystyle P_{1}}$ are given.

Remark 1: The 2-points–2-tangents property of a parabola is an affine version of the 3-point degeneration of Pascal's theorem.

Remark 2: The 2-points–2-tangents property should not be confused with the following property of a parabola, which also deals with 2 points and 2 tangents, but is not related to Pascal's theorem.

Axis direction

Construction of the axis direction

The statements above presume the knowledge of the axis direction of the parabola, in order to construct the points ${\displaystyle Q_{1},Q_{2}}$. The following property determines the points ${\displaystyle Q_{1},Q_{2}}$ by two given points and their tangents only, and the result is that the line ${\displaystyle Q_{1}Q_{2}}$ is parallel to the axis of the parabola.

Let

1. ${\displaystyle P_{1}=(x_{1},y_{1}),\ P_{2}=(x_{2},y_{2})}$ be two points of the parabola ${\displaystyle y=ax^{2}}$, and ${\displaystyle t_{1},t_{2}}$ be their tangents;
2. ${\displaystyle Q_{1}}$ be the intersection of the tangents ${\displaystyle t_{1},t_{2}}$,
3. ${\displaystyle Q_{2}}$ be the intersection of the parallel line to ${\displaystyle t_{1}}$ through ${\displaystyle P_{2}}$ with the parallel line to ${\displaystyle t_{2}}$ through ${\displaystyle P_{1}}$ (see picture).

Then the line ${\displaystyle Q_{1}Q_{2}}$ is parallel to the axis of the parabola and has the equation ${\displaystyle x=(x_{1}+x_{2})/2.}$

Proof: can be done (like the properties above) for the unit parabola ${\displaystyle y=x^{2}}$.

Application: This property can be used to determine the direction of the axis of a parabola, if two points and their tangents are given. An alternative way is to determine the midpoints of two parallel chords, see section on parallel chords.

Remark: This property is an affine version of the theorem of two perspective triangles of a non-degenerate conic.[10]

Inscribed angles of a parabola

A parabola with equation ${\displaystyle y=ax^{2}+bx+c,\ a\neq 0}$ is uniquely determined by three points ${\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),(x_{3},y_{3})}$ with different x coordinates. The usual procedure to determine the coefficients ${\displaystyle a,b,c}$ is to insert the point coordinates into the equation. The result is a linear system of three equations, which can be solved by Gaussian elimination or Cramer's rule, for example. An alternative way uses the inscribed angle theorem for parabolas.

In the following, the angle of two lines will be measured by the difference of the slopes of the line with respect to the directrix of the parabola. That is, for a parabola of equation ${\displaystyle y=ax^{2}+bx+c,}$ the angle between two lines of equations ${\displaystyle y=m_{1}x+d_{1},\ y=m_{2}x+d_{2}}$ is measured by ${\displaystyle m_{1}-m_{2}.}$

Analogous to the inscribed angle theorem for circles, one has the inscribed angle theorem for parabolas:[11][12]

Four points ${\displaystyle P_{i}=(x_{i},y_{i}),\ i=1,\ldots ,4,}$ with different x coordinates (see picture) are on a parabola with equation ${\displaystyle y=ax^{2}+bx+c}$ if and only if the angles at ${\displaystyle P_{3}}$ and ${\displaystyle P_{4}}$ have the same measure, as defined above. That is,

${\displaystyle {\frac {y_{4}-y_{1}}{x_{4}-x_{1}}}-{\frac {y_{4}-y_{2}}{x_{4}-x_{2}}}={\frac {y_{3}-y_{1}}{x_{3}-x_{1}}}-{\frac {y_{3}-y_{2}}{x_{3}-x_{2}}}.}$

(Proof: straightforward calculation: If the points are on a parabola, one may translate the coordinates for having the equation ${\displaystyle y=ax^{2}}$, then one has ${\displaystyle {\frac {y_{i}-y_{j}}{x_{i}-x_{j}}}=x_{i}+x_{j}}$ if the points are on the parabola.)

A consequence is that the equation (in ${\displaystyle {\color {green}x},{\color {red}y}}$) of the parabola determined by 3 points ${\displaystyle P_{i}=(x_{i},y_{i}),\ i=1,2,3,}$ with different x coordinates is (if two x coordinates are equal, there is no parabola with directrix parallel to the x axis, which passes through the points)

${\displaystyle {\frac {{\color {red}y}-y_{1}}{{\color {green}x}-x_{1}}}-{\frac {{\color {red}y}-y_{2}}{{\color {green}x}-x_{2}}}={\frac {y_{3}-y_{1}}{x_{3}-x_{1}}}-{\frac {y_{3}-y_{2}}{x_{3}-x_{2}}}.}$

Multiplying by the denominators that depend on ${\displaystyle {\color {green}x},}$ one obtains the more standard form

${\displaystyle (x_{1}-x_{2}){\color {red}y}=({\color {green}x}-x_{1})({\color {green}x}-x_{2})\left({\frac {y_{3}-y_{1}}{x_{3}-x_{1}}}-{\frac {y_{3}-y_{2}}{x_{3}-x_{2}}}\right)+(y_{1}-y_{2}){\color {green}x}+x_{1}y_{2}-x_{2}y_{1}.}$

Parabola: pole–polar relation

In a suitable coordinate system any parabola can be described by an equation ${\displaystyle y=ax^{2}}$. The equation of the tangent at a point ${\displaystyle P_{0}=(x_{0},y_{0}),\ y_{0}=ax_{0}^{2}}$ is

${\displaystyle y=2ax_{0}(x-x_{0})+y_{0}=2ax_{0}x-ax_{0}^{2}=2ax_{0}x-y_{0}.}$

One obtains the function

${\displaystyle (x_{0},y_{0})\to y=2ax_{0}x-y_{0}}$

on the set of points of the parabola onto the set of tangents.

Obviously, this function can be extended onto the set of all points of ${\displaystyle \mathbb {R} ^{2}}$ to a bijection between the points of ${\displaystyle \mathbb {R} ^{2}}$ and the lines with equations ${\displaystyle y=mx+d,\ m,d\in \mathbb {R} }$. The inverse mapping is

line ${\displaystyle y=mx+d}$ → point ${\displaystyle ({\tfrac {m}{2a}},-d)}$.

This relation is called the pole–polar relation of the parabola, where the point is the pole, and the corresponding line its polar.

By calculation, one checks the following properties of the pole–polar relation of the parabola:

• For a point (pole) on the parabola, the polar is the tangent at this point (see picture: ${\displaystyle P_{1},\ p_{1}}$).
• For a pole ${\displaystyle P}$ outside the parabola the intersection points of its polar with the parabola are the touching points of the two tangents passing ${\displaystyle P}$ (see picture: ${\displaystyle P_{2},\ p_{2}}$).
• For a point within the parabola the polar has no point with the parabola in common (see picture: ${\displaystyle P_{3},\ p_{3}}$ and ${\displaystyle P_{4},\ p_{4}}$).
• The intersection point of two polar lines (for example, ${\displaystyle p_{3},p_{4}}$) is the pole of the connecting line of their poles (in example: ${\displaystyle P_{3},P_{4}}$).
• Focus and directrix of the parabola are a pole–polar pair.

Remark: Pole–polar relations also exist for ellipses and hyperbolas.

S is the focus, and V is the principal vertex of the parabola VG. Draw VX perpendicular to SV.

Take any point B on VG and drop a perpendicular BQ from B to VX. Draw perpendicular ST intersecting BQ, extended if necessary, at T. At B draw the perpendicular BJ, intersecting VX at J.

For the parabola, the segment VBV, the area enclosed by the chord VB and the arc VB, is equal to ∆VBQ / 3, also ${\displaystyle BQ={\frac {VQ^{2}}{4SV}}}$.

The area of the parabolic sector SVB = ∆SVB + ∆VBQ / 3${\displaystyle {}={\frac {SV\cdot VQ}{2}}+{\frac {VQ\cdot BQ}{6}}}$.

Since triangles TSB and QBJ are similar,

${\displaystyle VJ=VQ-JQ=VQ-{\frac {BQ\cdot TB}{ST}}=VQ-{\frac {BQ\cdot (SV-BQ)}{VQ}}={\frac {3VQ}{4}}+{\frac {VQ\cdot BQ}{4SV}}.}$

Therefore, the area of the parabolic sector ${\displaystyle SVB={\frac {2SV\cdot VJ}{3}}}$ and can be found from the length of VJ, as found above.

A circle through S, V and B also passes through J.

Conversely, if a point, B on the parabola VG is to be found so that the area of the sector SVB is equal to a specified value, determine the point J on VX and construct a circle through S, V and J. Since SJ is the diameter, the center of the circle is at its midpoint, and it lies on the perpendicular bisector of SV, a distance of one half VJ from SV. The required point B is where this circle intersects the parabola.

If a body traces the path of the parabola due to an inverse square force directed towards S, the area SVB increases at a constant rate as point B moves forward. It follows that J moves at constant speed along VX as B moves along the parabola.

If the speed of the body at the vertex where it is moving perpendicularly to SV is v, then the speed of J is equal to 3v/4.

The construction can be extended simply to include the case where neither radius coincides with the axis SV as follows. Let A be a fixed point on VG between V and B, and point H be the intersection on VX with the perpendicular to SA at A. From the above, the area of the parabolic sector ${\displaystyle SAB={\frac {2SV\cdot (VJ-VH)}{3}}={\frac {2SV\cdot HJ}{3}}}$.

Conversely, if it is required to find the point B for a particular area SAB, find point J from HJ and point B as before. By Book 1, Proposition 16, Corollary 6 of Newton's Principia, the speed of a body moving along a parabola with a force directed towards the focus is inversely proportional to the square root of the radius. If the speed at A is v, then at the vertex V it is ${\displaystyle {\sqrt {\frac {SA}{SV}}}v}$, and point J moves at a constant speed of ${\displaystyle {\frac {3v}{4}}{\sqrt {\frac {SA}{SV}}}}$.

The above construction was devised by Isaac Newton and can be found in Book 1 of Philosophiæ Naturalis Principia Mathematica as Proposition 30.

The focal length of a parabola is half of its radius of curvature at its vertex.

Proof

• 
  Image is inverted. AB is x axis. C is origin. O is center. A is (x, y). OA = OC = R. PA = x. CP = y. OP = (R − y). Other points and lines are irrelevant for this purpose.
• 
  The radius of curvature at the vertex is twice the focal length. The measurements shown on the above diagram are in units of the latus rectum, which is four times the focal length.
• 

Consider a point (x, y) on a circle of radius R and with center at the point (0, R). The circle passes through the origin. If the point is near the origin, the Pythagorean theorem shows that

${\displaystyle {\begin{aligned}x^{2}+(R-y)^{2}&=R^{2},\\x^{2}+R^{2}-2Ry+y^{2}&=R^{2},\\x^{2}+y^{2}&=2Ry.\end{aligned}}}$

But if (x, y) is extremely close to the origin, since the x axis is a tangent to the circle, y is very small compared with x, so y² is negligible compared with the other terms. Therefore, extremely close to the origin

${\displaystyle x^{2}=2Ry.}$ (1)

Compare this with the parabola

${\displaystyle x^{2}=4fy,}$ (2)

which has its vertex at the origin, opens upward, and has focal length f (see preceding sections of this article).

Equations (1) and (2) are equivalent if R = 2f. Therefore, this is the condition for the circle and parabola to coincide at and extremely close to the origin. The radius of curvature at the origin, which is the vertex of the parabola, is twice the focal length.

Corollary

A concave mirror that is a small segment of a sphere behaves approximately like a parabolic mirror, focusing parallel light to a point midway between the centre and the surface of the sphere.

Parabola as an affine image of the unit parabola

Another definition of a parabola uses affine transformations:

• Any parabola is the affine image of the unit parabola with equation ${\displaystyle y=x^{2}}$.

parametric representation

An affine transformation of the Euclidean plane has the form ${\displaystyle {\vec {x}}\to {\vec {f}}_{0}+A{\vec {x}}}$, where ${\displaystyle A}$ is a regular matrix (determinant is not 0), and ${\displaystyle {\vec {f}}_{0}}$ is an arbitrary vector. If ${\displaystyle {\vec {f}}_{1},{\vec {f}}_{2}}$ are the column vectors of the matrix ${\displaystyle A}$, the unit parabola ${\displaystyle (t,t^{2}),\ t\in \mathbb {R} }$ is mapped onto the parabola

${\displaystyle {\vec {x}}={\vec {p}}(t)={\vec {f}}_{0}+{\vec {f}}_{1}t+{\vec {f}}_{2}t^{2},}$

where

${\displaystyle {\vec {f}}_{0}}$ is a point of the parabola,

${\displaystyle {\vec {f}}_{1}}$ is a tangent vector at point ${\displaystyle {\vec {f}}_{0}}$,

${\displaystyle {\vec {f}}_{2}}$ is parallel to the axis of the parabola (axis of symmetry through the vertex).

vertex

In general, the two vectors ${\displaystyle {\vec {f}}_{1},{\vec {f}}_{2}}$ are not perpendicular, and ${\displaystyle {\vec {f}}_{0}}$ is not the vertex, unless the affine transformation is a similarity.

The tangent vector at the point ${\displaystyle {\vec {p}}(t)}$ is ${\displaystyle {\vec {p}}'(t)={\vec {f}}_{1}+2t{\vec {f}}_{2}}$. At the vertex the tangent vector is orthogonal to ${\displaystyle {\vec {f}}_{2}}$. Hence the parameter ${\displaystyle t_{0}}$ of the vertex is the solution of the equation

${\displaystyle {\vec {p}}'(t)\cdot {\vec {f}}_{2}={\vec {f}}_{1}\cdot {\vec {f}}_{2}+2tf_{2}^{2}=0,}$

which is

${\displaystyle t_{0}=-{\frac {{\vec {f}}_{1}\cdot {\vec {f}}_{2}}{2f_{2}^{2}}},}$

and the vertex is

${\displaystyle {\vec {p}}(t_{0})={\vec {f}}_{0}-{\frac {{\vec {f}}_{1}\cdot {\vec {f}}_{2}}{2f_{2}^{2}}}{\vec {f}}_{1}+{\frac {({\vec {f}}_{1}\cdot {\vec {f}}_{2})^{2}}{4(f_{2}^{2})^{2}}}{\vec {f}}_{2}.}$

focal length and focus

The focal length can be determined by a suitable parameter transformation (which does not change the geometric shape of the parabola). The focal length is

${\displaystyle f={\frac {f_{1}^{2}\,f_{2}^{2}-({\vec {f}}_{1}\cdot {\vec {f}}_{2})^{2}}{4|f_{2}|^{3}}}.}$

Hence the focus of the parabola is

${\displaystyle F:\ {\vec {f}}_{0}-{\frac {{\vec {f}}_{1}\cdot {\vec {f}}_{2}}{2f_{2}^{2}}}{\vec {f}}_{1}+{\frac {f_{1}^{2}\,f_{2}^{2}}{4(f_{2}^{2})^{2}}}{\vec {f}}_{2}.}$

implicit representation

Solving the parametric representation for ${\displaystyle \;t,t^{2}\;}$ by Cramer's rule and using ${\displaystyle \;t\cdot t-t^{2}=0\;}$, one gets the implicit representation

${\displaystyle \det({\vec {x}}\!-\!{\vec {f}}\!_{0},{\vec {f}}\!_{2})^{2}-\det({\vec {f}}\!_{1},{\vec {x}}\!-\!{\vec {f}}\!_{0})\det({\vec {f}}\!_{1},{\vec {f}}\!_{2})=0}$.

parabola in space

The definition of a parabola in this section gives a parametric representation of an arbitrary parabola, even in space, if one allows ${\displaystyle {\vec {f}}\!_{0},{\vec {f}}\!_{1},{\vec {f}}\!_{2}}$ to be vectors in space.

Quadratic Bézier curve and its control points

A quadratic Bézier curve is a curve ${\displaystyle {\vec {c}}(t)}$ defined by three points ${\displaystyle P_{0}:{\vec {p}}_{0}}$, ${\displaystyle P_{1}:{\vec {p}}_{1}}$ and ${\displaystyle P_{2}:{\vec {p}}_{2}}$, called its control points: