Axis of symmetry parallel to the y axis

Parabola with axis parallel to y-axis; p is the semi-latus rectum

If one introduces Cartesian coordinates, such that ${\displaystyle F=(0,f),\ f>0,}$ and the directrix has the equation ${\displaystyle y=-f}$, one obtains for a point ${\displaystyle P=(x,y)}$ from ${\displaystyle |PF|^{2}=|Pl|^{2}}$ the equation ${\displaystyle x^{2}+(y-f)^{2}=(y+f)^{2}}$. Solving for ${\displaystyle y}$ yields

$${\displaystyle y={\frac {1}{4f}}x^{2}.}$$

This parabola is U-shaped (opening to the top).

The horizontal chord through the focus (see picture in opening section) is called the latus rectum; one half of it is the semi-latus rectum. The latus rectum is parallel to the directrix. The semi-latus rectum is designated by the letter ${\displaystyle p}$. From the picture one obtains

$${\displaystyle p=2f.}$$

The latus rectum is defined similarly for the other two conics – the ellipse and the hyperbola. The latus rectum is the line drawn through a focus of a conic section parallel to the directrix and terminated both ways by the curve. For any case, ${\displaystyle p}$ is the radius of the osculating circle at the vertex. For a parabola, the semi-latus rectum, ${\displaystyle p}$, is the distance of the focus from the directrix. Using the parameter ${\displaystyle p}$, the equation of the parabola can be rewritten as

$${\displaystyle x^{2}=2py.}$$

More generally, if the vertex is ${\displaystyle V=(v_{1},v_{2})}$, the focus ${\displaystyle F=(v_{1},v_{2}+f)}$, and the directrix ${\displaystyle y=v_{2}-f}$, one obtains the equation

$${\displaystyle y={\frac {1}{4f}}(x-v_{1})^{2}+v_{2}={\frac {1}{4f}}x^{2}-{\frac {v_{1}}{2f}}x+{\frac {v_{1}^{2}}{4f}}+v_{2}.}$$

Remarks:

1. In the case of ${\displaystyle f<0}$ the parabola has a downward opening.
2. The presumption that the axis is parallel to the y axis allows one to consider a parabola as the graph of a polynomial of degree 2, and conversely: the graph of an arbitrary polynomial of degree 2 is a parabola (see next section).
3. If one exchanges ${\displaystyle x}$ and ${\displaystyle y}$, one obtains equations of the form ${\displaystyle y^{2}=2px}$. These parabolas open to the left (if ${\displaystyle p<0}$) or to the right (if ${\displaystyle p>0}$).

General position

Parabola: general position

If the focus is ${\displaystyle F=(f_{1},f_{2})}$, and the directrix ${\displaystyle ax+by+c=0}$, then one obtains the equation

$${\displaystyle {\frac {(ax+by+c)^{2}}{a^{2}+b^{2}}}=(x-f_{1})^{2}+(y-f_{2})^{2}}$$

(the left side of the equation uses the Hesse normal form of a line to calculate the distance ${\displaystyle |Pl|}$).

For a parametric equation of a parabola in general position see § As the affine image of the unit parabola.

The implicit equation of a parabola is defined by an irreducible polynomial of degree two:

$${\displaystyle ax^{2}+bxy+cy^{2}+dx+ey+f=0,}$$

such that ${\displaystyle b^{2}-4ac=0,}$ or, equivalently, such that ${\displaystyle ax^{2}+bxy+cy^{2}}$ is the square of a linear polynomial.

Diagram, description, and definitions

Cone with cross-sections

The diagram represents a cone with its axis AV. The point A is its apex. An inclined cross-section of the cone, shown in pink, is inclined from the axis 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.

A cross-section perpendicular to the axis of the cone passes through the vertex P of the parabola. This cross-section is circular, but appears elliptical when viewed obliquely, as is shown in the diagram. Its centre is V, and PK is a diameter. We will call its radius r.

Another perpendicular to the axis, circular cross-section of the cone is farther from the apex A than the one just described. It has a chord DE, which joins the points where the parabola intersects the circle. Another chord BC is the perpendicular bisector of DE and is consequently a diameter of the circle. These two chords and the parabola's axis of symmetry PM all intersect at the point M.

All the labelled points, except D and E, are coplanar. They are in the plane of symmetry of the whole figure. This includes the point F, which is not mentioned above. It is defined and discussed below, in § Position of the focus.

Let us call the length of DM and of EM x, and the length of PM y.

Derivation of quadratic equation

The lengths of BM and CM are:

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

$${\displaystyle {\overline {\mathrm {BM} }}\cdot {\overline {\mathrm {CM} }}={\overline {\mathrm {DM} }}\cdot {\overline {\mathrm {EM} }}.}$$

Substituting:

$${\displaystyle 4ry\sin \theta =x^{2}.}$$

Rearranging:

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

For any given cone and parabola, r and θ are constants, but x and y are variables that depend on the arbitrary height at which the horizontal cross-section BECD is made. This last equation shows the relationship between these variables. They can be interpreted as Cartesian coordinates of the points D and E, in a system in the pink plane with P as its origin. Since x is squared in the equation, the fact that D and E are on opposite sides of the y axis is unimportant. If the horizontal cross-section moves up or down, toward or away from the apex of the cone, D and E move along the parabola, always maintaining the relationship between x and y shown in the equation. The parabolic curve is therefore the locus of points where the equation is satisfied, which makes it a Cartesian graph of the quadratic function in the equation.

Focal length

It is proved in a preceding section that if a parabola has its vertex at the origin, and if it opens in the positive y direction, then its equation is y = x²/4f, where f is its focal length.^{[lower-alpha 2]} Comparing this with the last equation above shows that the focal length of the parabola in the cone is r sin θ.

Position of the focus

In the diagram above, the point V is the foot of the perpendicular from the vertex of the parabola to the axis of the cone. The point F is the foot of the perpendicular from the point V to the plane of the parabola.^{[lower-alpha 3]} By symmetry, F is on the axis of symmetry of the parabola. 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, the distance of F from the vertex of the parabola is r sin θ. It is shown above that this distance equals the focal length of the parabola, which is the distance from the vertex to the focus. The focus and the point F are therefore equally distant from the vertex, along the same line, which implies that they are the same point. Therefore, the point F, defined above, is the focus of the parabola.

This discussion started from the definition of a parabola as a conic section, but it has now led to a description as a graph of a quadratic function. This shows that these two descriptions are equivalent. They both define curves of exactly the same shape.

Alternative proof with Dandelin spheres

Parabola (red): side projection view and top projection view of a cone with a Dandelin sphere

An alternative proof can be done using Dandelin spheres. It works without calculation and uses elementary geometric considerations only (see the derivation below).

The intersection of an upright cone by a plane ${\displaystyle \pi }$, whose inclination from vertical is the same as a generatrix (a.k.a. generator line, a line containing the apex and a point on the cone surface) ${\displaystyle m_{0}}$ of the cone, is a parabola (red curve in the diagram).

This generatrix ${\displaystyle m_{0}}$ is the only generatrix of the cone that is parallel to plane ${\displaystyle \pi }$. Otherwise, if there are two generatrices parallel to the intersecting plane, the intersection curve will be a hyperbola (or degenerate hyperbola, if the two generatrices are in the intersecting plane). If there is no generatrix parallel to the intersecting plane, the intersection curve will be an ellipse or a circle (or a point).

Let plane ${\displaystyle \sigma }$ be the plane that contains the vertical axis of the cone and line ${\displaystyle m_{0}}$. The inclination of plane ${\displaystyle \pi }$ from vertical is the same as line ${\displaystyle m_{0}}$ means that, viewing from the side (that is, the plane ${\displaystyle \pi }$ is perpendicular to plane ${\displaystyle \sigma }$), ${\displaystyle m_{0}\parallel \pi }$.

In order to prove the directrix property of a parabola (see § Definition as a locus of points above), one uses a Dandelin sphere ${\displaystyle d}$, which is a sphere that touches the cone along a circle ${\displaystyle c}$ and plane ${\displaystyle \pi }$ at point ${\displaystyle F}$. The plane containing the circle ${\displaystyle c}$ intersects with plane ${\displaystyle \pi }$ at line ${\displaystyle l}$. There is a mirror symmetry in the system consisting of plane ${\displaystyle \pi }$, Dandelin sphere ${\displaystyle d}$ and the cone (the plane of symmetry is ${\displaystyle \sigma }$).

Since the plane containing the circle ${\displaystyle c}$ is perpendicular to plane ${\displaystyle \sigma }$, and ${\displaystyle \pi \perp \sigma }$, their intersection line ${\displaystyle l}$ must also be perpendicular to plane ${\displaystyle \sigma }$. Since line ${\displaystyle m_{0}}$ is in plane ${\displaystyle \sigma }$, ${\displaystyle l\perp m_{0}}$.

It turns out that ${\displaystyle F}$ is the focus of the parabola, and ${\displaystyle l}$ is the directrix of the parabola.

1. Let ${\displaystyle P}$ be an arbitrary point of the intersection curve.
2. The generatrix of the cone containing ${\displaystyle P}$ intersects circle ${\displaystyle c}$ at point ${\displaystyle A}$.
3. The line segments ${\displaystyle {\overline {PF}}}$ and ${\displaystyle {\overline {PA}}}$ are tangential to the sphere ${\displaystyle d}$, and hence are of equal length.
4. Generatrix ${\displaystyle m_{0}}$ intersects the circle ${\displaystyle c}$ at point ${\displaystyle D}$. The line segments ${\displaystyle {\overline {ZD}}}$ and ${\displaystyle {\overline {ZA}}}$ are tangential to the sphere ${\displaystyle d}$, and hence are of equal length.
5. Let line ${\displaystyle q}$ be the line parallel to ${\displaystyle m_{0}}$ and passing through point ${\displaystyle P}$. Since ${\displaystyle m_{0}\parallel \pi }$, and point ${\displaystyle P}$ is in plane ${\displaystyle \pi }$, line ${\displaystyle q}$ must be in plane ${\displaystyle \pi }$. Since ${\displaystyle m_{0}\perp l}$, we know that ${\displaystyle q\perp l}$ as well.
6. Let point ${\displaystyle B}$ be the foot of the perpendicular from point ${\displaystyle P}$ to line ${\displaystyle l}$, that is, ${\displaystyle {\overline {PB}}}$ is a segment of line ${\displaystyle q}$, and hence ${\displaystyle {\overline {PB}}\parallel {\overline {ZD}}}$.
7. From intercept theorem and ${\displaystyle {\overline {ZD}}={\overline {ZA}}}$ we know that ${\displaystyle {\overline {PA}}={\overline {PB}}}$. Since ${\displaystyle {\overline {PA}}={\overline {PF}}}$, we know that ${\displaystyle {\overline {PF}}={\overline {PB}}}$, which means that the distance from ${\displaystyle P}$ to the focus ${\displaystyle F}$ is equal to the distance from ${\displaystyle P}$ to the directrix ${\displaystyle l}$.

Construction and definitions

The point E is an arbitrary point on the parabola. The focus is F, the vertex is A (the origin), and the line FA is the axis of symmetry. The line EC is parallel to the axis of symmetry, intersects the x axis at D and intersects the directrix at C. The point B is the midpoint of the line segment FC.

Deductions

The vertex A is equidistant from the focus F and from the directrix. 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. Its x coordinate is half that of D, that is, x/2. The slope of the line BE is the quotient of the lengths of ED and BD, which is x²/x/2 = 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 α are congruent. (The angle above E is vertically opposite angle ∠BEC.) This means that a ray of light that 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 infinitesimal 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.

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.

Other consequences

There are other theorems that can be deduced simply from the above argument.

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.

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]

Parabola

Steiner generation of a parabola

Steiner established the following procedure for the construction of a non-degenerate conic (see Steiner conic):

Given two pencils ${\displaystyle B(U),B(V)}$ of lines at two points ${\displaystyle U,V}$ (all lines containing ${\displaystyle U}$ and ${\displaystyle V}$ respectively) and a projective but not perspective mapping ${\displaystyle \pi }$ of ${\displaystyle B(U)}$ onto ${\displaystyle B(V)}$, the intersection points of corresponding lines form a non-degenerate projective conic section.

This procedure can be used for a simple construction of points on the parabola ${\displaystyle y=ax^{2}}$:

• Consider the pencil at the vertex ${\displaystyle S(0,0)}$ and the set of lines ${\displaystyle \Pi _{y}}$ that are parallel to the y axis.
  1. Let ${\displaystyle P=(x_{0},y_{0})}$ be a point on the parabola, and ${\displaystyle A=(0,y_{0})}$, ${\displaystyle B=(x_{0},0)}$.
  2. The line segment ${\displaystyle {\overline {BP}}}$ is divided into n equally spaced segments, and this division is projected (in the direction ${\displaystyle BA}$) onto the line segment ${\displaystyle {\overline {AP}}}$ (see figure). This projection gives rise to a projective mapping ${\displaystyle \pi }$ from pencil ${\displaystyle S}$ onto the pencil ${\displaystyle \Pi _{y}}$.
  3. The intersection of the line ${\displaystyle SB_{i}}$ and the i-th parallel to the y axis is a point on the parabola.

Proof: straightforward calculation.

Remark: Steiner's generation is also available for ellipses and hyperbolas.

Dual parabola

Dual parabola and Bezier curve of degree 2 (right: curve point and division points ${\displaystyle Q_{0},Q_{1}}$ for parameter ${\displaystyle t=0.4}$)

A dual parabola consists of the set of tangents of an ordinary parabola.

The Steiner generation of a conic can be applied to the generation of a dual conic by changing the meanings of points and lines:

Let be given two point sets on two lines ${\displaystyle u,v}$, and a projective but not perspective mapping ${\displaystyle \pi }$ between these point sets, then the connecting lines of corresponding points form a non degenerate dual conic.

In order to generate elements of a dual parabola, one starts with

1. three points ${\displaystyle P_{0},P_{1},P_{2}}$ not on a line,
2. divides the line sections ${\displaystyle {\overline {P_{0}P_{1}}}}$ and ${\displaystyle {\overline {P_{1}P_{2}}}}$ each into ${\displaystyle n}$ equally spaced line segments and adds numbers as shown in the picture.
3. Then the lines ${\displaystyle P_{0}P_{1},P_{1}P_{2},(1,1),(2,2),\dotsc }$ are tangents of a parabola, hence elements of a dual parabola.
4. The parabola is a Bezier curve of degree 2 with the control points ${\displaystyle P_{0},P_{1},P_{2}}$.

The proof is a consequence of the de Casteljau algorithm for a Bezier curve of degree 2.

Two tangent properties related to the latus rectum

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.^[13]: 26

Perpendicular tangents intersect on the directrix

Orthoptic property

If two tangents to a parabola are perpendicular to each other, then they intersect on the directrix. Conversely, two tangents that intersect on the directrix are perpendicular. In other words, at any point on the directrix the whole parabola subtends a right angle.

Lambert's theorem

Let three tangents to a parabola form a triangle. Then Lambert's theorem states that the focus of the parabola lies on the circumcircle of the triangle.^{[14][8]: Corollary 20}

Tsukerman's converse to Lambert's theorem states that, given three lines that bound a triangle, if two of the lines are tangent to a parabola whose focus lies on the circumcircle of the triangle, then the third line is also tangent to the parabola.^[15]

Focal length calculated from parameters of a chord

Suppose a chord crosses a parabola perpendicular to its axis of symmetry. Let the length of the chord between the points where it intersects the parabola be c and the distance from the vertex of the parabola to the chord, measured along the axis of symmetry, be d. The focal length, f, of the parabola is given by

$${\displaystyle f={\frac {c^{2}}{16d}}.}$$

Area enclosed between a parabola and a chord

Parabola (magenta) and line (lower light blue) including a chord (blue). The area enclosed between them is in pink. The chord itself ends at the points where the line intersects the parabola.

The area enclosed between a parabola and a chord (see diagram) is two-thirds of the area of a parallelogram that surrounds it. One side of the parallelogram is the chord, and the opposite side is a tangent to the parabola.^[16][17] The slope of the other parallel sides is irrelevant to the area. Often, as here, they are drawn parallel with the parabola's axis of symmetry, but this is arbitrary.

A theorem equivalent to this one, but different in details, was derived by Archimedes in the 3rd century BCE. He used the areas of triangles, rather than that of the parallelogram.^{[lower-alpha 4]} See The Quadrature of the Parabola.

If the chord has length b and is perpendicular to the parabola's axis of symmetry, and if the perpendicular distance from the parabola's vertex to the chord is h, the parallelogram is a rectangle, with sides of b and h. The area A of the parabolic segment enclosed by the parabola and the chord is therefore

$${\displaystyle A={\frac {2}{3}}bh.}$$

This formula can be compared with the area of a triangle: 1/2bh.

In general, the enclosed area can be calculated as follows. First, locate the point on the parabola where its slope equals that of the chord. This can be done with calculus, or by using a line that is parallel to the axis of symmetry of the parabola and passes through the midpoint of the chord. The required point is where this line intersects the parabola.^{[lower-alpha 5]} Then, using the formula given in Distance from a point to a line, calculate the perpendicular distance from this point to the chord. Multiply this by the length of the chord to get the area of the parallelogram, then by 2/3 to get the required enclosed area.

Corollary concerning midpoints and endpoints of chords

Midpoints of parallel chords

A corollary of the above discussion is that if a parabola has several parallel chords, their midpoints all lie on a line parallel to the axis of symmetry. If tangents to the parabola are drawn through the endpoints of any of these chords, the two tangents intersect on this same line parallel to the axis of symmetry (see Axis-direction of a parabola).^{[lower-alpha 6]}

Arc length

If a point X is located on a parabola with 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 that terminate at X can be calculated from f and p as follows, assuming they are all expressed in the same units.^{[lower-alpha 7]}

$${\displaystyle {\begin{aligned}h&={\frac {p}{2}},\\q&={\sqrt {f^{2}+h^{2}}},\\s&={\frac {hq}{f}}+f\ln {\frac {h+q}{f}}.\end{aligned}}}$$

This quantity s is the length of the arc between X and the vertex of the parabola.

The length of the arc between X and the symmetrically opposite point on the other side of the parabola is 2s.

The perpendicular distance p can be given a positive or negative sign to indicate on which side of the axis of symmetry X is situated. Reversing the sign of p reverses the signs of h and s without changing their absolute values. If these quantities are signed, the length of the arc between any two points on the parabola is always shown by the difference between their values of s. The calculation can be simplified by using the properties of logarithms:

$${\displaystyle s_{1}-s_{2}={\frac {h_{1}q_{1}-h_{2}q_{2}}{f}}+f\ln {\frac {h_{1}+q_{1}}{h_{2}+q_{2}}}.}$$

This can be useful, for example, in calculating the size of the material needed to make a parabolic reflector or parabolic trough.

This calculation can be used for a parabola in any orientation. It is not restricted to the situation where the axis of symmetry is parallel to the y axis.

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.