Ellipse

The standard form of an ellipse in Cartesian coordinates assumes that the origin is the center of the ellipse, the x-axis is the major axis, and:

For an arbitrary point ${\displaystyle (x,y)}$ the distance to the focus ${\displaystyle (c,0)}$ is ${\textstyle {\sqrt {(x-c)^{2}+y^{2}}}}$ and to the other focus ${\textstyle {\sqrt {(x+c)^{2}+y^{2}}}}$. Hence the point ${\displaystyle (x,\,y)}$ is on the ellipse whenever:

$${\displaystyle {\sqrt {(x-c)^{2}+y^{2}}}+{\sqrt {(x+c)^{2}+y^{2}}}=2a\ .}$$

Removing the radicals by suitable squarings and using ${\displaystyle b^{2}=a^{2}-c^{2}}$ (see diagram) produces the standard equation of the ellipse:[3]

$${\displaystyle {\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=1,}$$

or, solved for y:

$${\displaystyle y=\pm {\frac {b}{a}}{\sqrt {a^{2}-x^{2}}}=\pm {\sqrt {\left(a^{2}-x^{2}\right)\left(1-e^{2}\right)}}.}$$

The width and height parameters ${\displaystyle a,\;b}$ are called the semi-major and semi-minor axes. The top and bottom points ${\displaystyle V_{3}=(0,\,b),\;V_{4}=(0,\,-b)}$ are the co-vertices. The distances from a point ${\displaystyle (x,\,y)}$ on the ellipse to the left and right foci are ${\displaystyle a+ex}$ and ${\displaystyle a-ex}$.

It follows from the equation that the ellipse is symmetric with respect to the coordinate axes and hence with respect to the origin.

Throughout this article, the semi-major and semi-minor axes are denoted ${\displaystyle a}$ and ${\displaystyle b}$, respectively, i.e. ${\displaystyle a\geq b>0\ .}$

In principle, the canonical ellipse equation ${\displaystyle {\tfrac {x^{2}}{a^{2}}}+{\tfrac {y^{2}}{b^{2}}}=1}$ may have ${\displaystyle a<b}$ (and hence the ellipse would be taller than it is wide). This form can be converted to the standard form by transposing the variable names ${\displaystyle x}$ and ${\displaystyle y}$ and the parameter names ${\displaystyle a}$ and ${\displaystyle b.}$

This is the distance from the center to a focus: ${\displaystyle c={\sqrt {a^{2}-b^{2}}}}$.

The eccentricity can be expressed as:

$${\displaystyle e={\frac {c}{a}}={\sqrt {1-\left({\frac {b}{a}}\right)^{2}}},}$$

assuming ${\displaystyle a>b.}$ An ellipse with equal axes (${\displaystyle a=b}$) has zero eccentricity, and is a circle.

The length of the chord through one focus, perpendicular to the major axis, is called the latus rectum. One half of it is the semi-latus rectum ${\displaystyle \ell }$. A calculation shows:[4]

$${\displaystyle \ell ={\frac {b^{2}}{a}}=a\left(1-e^{2}\right).}$$

The semi-latus rectum ${\displaystyle \ell }$ is equal to the radius of curvature at the vertices (see section curvature).

An arbitrary line ${\displaystyle g}$ intersects an ellipse at 0, 1, or 2 points, respectively called an exterior line, tangent and secant. Through any point of an ellipse there is a unique tangent. The tangent at a point ${\displaystyle (x_{1},\,y_{1})}$ of the ellipse ${\displaystyle {\tfrac {x^{2}}{a^{2}}}+{\tfrac {y^{2}}{b^{2}}}=1}$ has the coordinate equation:

$${\displaystyle {\frac {x_{1}}{a^{2}}}x+{\frac {y_{1}}{b^{2}}}y=1.}$$

A vector parametric equation of the tangent is:

$${\displaystyle {\vec {x}}={\begin{pmatrix}x_{1}\\y_{1}\end{pmatrix}}+s\left({\begin{array}{r}-y_{1}a^{2}\\x_{1}b^{2}\end{array}}\right),\quad s\in \mathbb {R} .}$$

Proof:

Let ${\displaystyle (x_{1},\,y_{1})}$ be a point on an ellipse and ${\textstyle {\vec {x}}={\begin{pmatrix}x_{1}\\y_{1}\end{pmatrix}}+s{\begin{pmatrix}u\\v\end{pmatrix}}}$ be the equation of any line ${\displaystyle g}$ containing ${\displaystyle (x_{1},\,y_{1})}$. Inserting the line's equation into the ellipse equation and respecting ${\textstyle {\frac {x_{1}^{2}}{a^{2}}}+{\frac {y_{1}^{2}}{b^{2}}}=1}$ yields:

$${\displaystyle {\frac {\left(x_{1}+su\right)^{2}}{a^{2}}}+{\frac {\left(y_{1}+sv\right)^{2}}{b^{2}}}=1\ \quad \Longrightarrow \quad 2s\left({\frac {x_{1}u}{a^{2}}}+{\frac {y_{1}v}{b^{2}}}\right)+s^{2}\left({\frac {u^{2}}{a^{2}}}+{\frac {v^{2}}{b^{2}}}\right)=0\ .}$$

There are then cases:

1. ${\displaystyle {\frac {x_{1}}{a^{2}}}u+{\frac {y_{1}}{b^{2}}}v=0.}$ Then line ${\displaystyle g}$ and the ellipse have only point ${\displaystyle (x_{1},\,y_{1})}$ in common, and ${\displaystyle g}$ is a tangent. The tangent direction has perpendicular vector ${\displaystyle {\begin{pmatrix}{\frac {x_{1}}{a^{2}}}&{\frac {y_{1}}{b^{2}}}\end{pmatrix}}}$, so the tangent line has equation ${\textstyle {\frac {x_{1}}{a^{2}}}x+{\tfrac {y_{1}}{b^{2}}}y=k}$ for some ${\displaystyle k}$. Because ${\displaystyle (x_{1},\,y_{1})}$ is on the tangent and the ellipse, one obtains ${\displaystyle k=1}$.
2. ${\displaystyle {\frac {x_{1}}{a^{2}}}u+{\frac {y_{1}}{b^{2}}}v\neq 0.}$ Then line ${\displaystyle g}$ has a second point in common with the ellipse, and is a secant.

Using (1) one finds that ${\displaystyle {\begin{pmatrix}-y_{1}a^{2}&x_{1}b^{2}\end{pmatrix}}}$ is a tangent vector at point ${\displaystyle (x_{1},\,y_{1})}$, which proves the vector equation.

If ${\displaystyle (x_{1},y_{1})}$ and ${\displaystyle (u,v)}$ are two points of the ellipse such that ${\textstyle {\frac {x_{1}u}{a^{2}}}+{\tfrac {y_{1}v}{b^{2}}}=0}$, then the points lie on two conjugate diameters (see below). (If ${\displaystyle a=b}$, the ellipse is a circle and "conjugate" means "orthogonal".)

If the standard ellipse is shifted to have center ${\displaystyle \left(x_{\circ },\,y_{\circ }\right)}$, its equation is

$${\displaystyle {\frac {\left(x-x_{\circ }\right)^{2}}{a^{2}}}+{\frac {\left(y-y_{\circ }\right)^{2}}{b^{2}}}=1\ .}$$

The axes are still parallel to the x- and y-axes.

In analytic geometry, the ellipse is defined as a quadric: the set of points ${\displaystyle (X,\,Y)}$ of the Cartesian plane that, in non-degenerate cases, satisfy the implicit equation[5][6]

$${\displaystyle AX^{2}+BXY+CY^{2}+DX+EY+F=0}$$

provided ${\displaystyle B^{2}-4AC<0.}$

To distinguish the degenerate cases from the non-degenerate case, let ∆ be the determinant

$${\displaystyle \Delta ={\begin{vmatrix}A&{\frac {1}{2}}B&{\frac {1}{2}}D\\{\frac {1}{2}}B&C&{\frac {1}{2}}E\\{\frac {1}{2}}D&{\frac {1}{2}}E&F\end{vmatrix}}=\left(AC-{\tfrac {1}{4}}B^{2}\right)F+{\tfrac {1}{4}}BED-{\tfrac {1}{4}}CD^{2}-{\tfrac {1}{4}}AE^{2}.}$$

Then the ellipse is a non-degenerate real ellipse if and only if C∆ < 0. If C∆ > 0, we have an imaginary ellipse, and if ∆ = 0, we have a point ellipse.[7]^: 63

The general equation's coefficients can be obtained from known semi-major axis ${\displaystyle a}$, semi-minor axis ${\displaystyle b}$, center coordinates ${\displaystyle \left(x_{\circ },\,y_{\circ }\right)}$, and rotation angle ${\displaystyle \theta }$ (the angle from the positive horizontal axis to the ellipse's major axis) using the formulae:

$${\displaystyle {\begin{aligned}A&=a^{2}\sin ^{2}\theta +b^{2}\cos ^{2}\theta &B&=2\left(b^{2}-a^{2}\right)\sin \theta \cos \theta \\[1ex]C&=a^{2}\cos ^{2}\theta +b^{2}\sin ^{2}\theta &D&=-2Ax_{\circ }-By_{\circ }\\[1ex]E&=-Bx_{\circ }-2Cy_{\circ }&F&=Ax_{\circ }^{2}+Bx_{\circ }y_{\circ }+Cy_{\circ }^{2}-a^{2}b^{2}.\end{aligned}}}$$

These expressions can be derived from the canonical equation

$${\displaystyle {\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=1}$$

by an affine transformation of the coordinates ${\displaystyle (x,\,y)}$:

$${\displaystyle {\begin{aligned}x&=\left(X-x_{\circ }\right)\cos \theta +\left(Y-y_{\circ }\right)\sin \theta \\y&=-\left(X-x_{\circ }\right)\sin \theta +\left(Y-y_{\circ }\right)\cos \theta .\end{aligned}}}$$

Conversely, the canonical form parameters can be obtained from the general form coefficients by the equations:

$${\displaystyle {\begin{aligned}a,b&={\frac {-{\sqrt {2{\big (}AE^{2}+CD^{2}-BDE+(B^{2}-4AC)F{\big )}{\big (}(A+C)\pm {\sqrt {(A-C)^{2}+B^{2}}}{\big )}}}}{B^{2}-4AC}}\\x_{\circ }&={\frac {2CD-BE}{B^{2}-4AC}}\\[5mu]y_{\circ }&={\frac {2AE-BD}{B^{2}-4AC}}\\[5mu]\theta &={\begin{cases}\operatorname {arccot} {\dfrac {C-A-{\sqrt {(A-C)^{2}+B^{2}}}}{B}}&{\text{for }}B\neq 0\\[5mu]0&{\text{for }}B=0,\ A<C\\[10mu]90^{\circ }&{\text{for }}B=0,\ A>C\\\end{cases}}\end{aligned}}}$$

Using trigonometric functions, a parametric representation of the standard ellipse ${\displaystyle {\tfrac {x^{2}}{a^{2}}}+{\tfrac {y^{2}}{b^{2}}}=1}$ is:

$${\displaystyle (x,\,y)=(a\cos t,\,b\sin t),\ 0\leq t<2\pi \,.}$$

The parameter t (called the eccentric anomaly in astronomy) is not the angle of ${\displaystyle (x(t),y(t))}$ with the x-axis, but has a geometric meaning due to Philippe de La Hire (see § Drawing ellipses below).[8]

With the substitution ${\textstyle u=\tan \left({\frac {t}{2}}\right)}$ and trigonometric formulae one obtains

$${\displaystyle \cos t={\frac {1-u^{2}}{1+u^{2}}}\ ,\quad \sin t={\frac {2u}{1+u^{2}}}}$$

and the rational parametric equation of an ellipse

$${\displaystyle {\begin{cases}x(u)=a\,{\dfrac {1-u^{2}}{1+u^{2}}}\\[10mu]y(u)=b\,{\dfrac {2u}{1+u^{2}}}\\[10mu]-\infty <u<\infty \end{cases}}}$$

which covers any point of the ellipse ${\displaystyle {\tfrac {x^{2}}{a^{2}}}+{\tfrac {y^{2}}{b^{2}}}=1}$ except the left vertex ${\displaystyle (-a,\,0)}$.

For ${\displaystyle u\in [0,\,1],}$ this formula represents the right upper quarter of the ellipse moving counter-clockwise with increasing ${\displaystyle u.}$ The left vertex is the limit ${\textstyle \lim _{u\to \pm \infty }(x(u),\,y(u))=(-a,\,0)\;.}$

Alternately, if the parameter ${\displaystyle [u:v]}$ is considered to be a point on the real projective line ${\textstyle \mathbf {P} (\mathbf {R} )}$, then the corresponding rational parametrization is

$${\displaystyle [u:v]\mapsto \left(a{\frac {v^{2}-u^{2}}{v^{2}+u^{2}}},b{\frac {2uv}{v^{2}+u^{2}}}\right).}$$

Then ${\textstyle [1:0]\mapsto (-a,\,0).}$

Rational representations of conic sections are commonly used in computer-aided design (see Bezier curve).

A parametric representation, which uses the slope ${\displaystyle m}$ of the tangent at a point of the ellipse can be obtained from the derivative of the standard representation ${\displaystyle {\vec {x}}(t)=(a\cos t,\,b\sin t)^{\mathsf {T}}}$:

$${\displaystyle {\vec {x}}'(t)=(-a\sin t,\,b\cos t)^{\mathsf {T}}\quad \rightarrow \quad m=-{\frac {b}{a}}\cot t\quad \rightarrow \quad \cot t=-{\frac {ma}{b}}.}$$

With help of trigonometric formulae one obtains:

$${\displaystyle \cos t={\frac {\cot t}{\pm {\sqrt {1+\cot ^{2}t}}}}={\frac {-ma}{\pm {\sqrt {m^{2}a^{2}+b^{2}}}}}\ ,\quad \quad \sin t={\frac {1}{\pm {\sqrt {1+\cot ^{2}t}}}}={\frac {b}{\pm {\sqrt {m^{2}a^{2}+b^{2}}}}}.}$$

Replacing ${\displaystyle \cos t}$ and ${\displaystyle \sin t}$ of the standard representation yields:

$${\displaystyle {\vec {c}}_{\pm }(m)=\left(-{\frac {ma^{2}}{\pm {\sqrt {m^{2}a^{2}+b^{2}}}}},\;{\frac {b^{2}}{\pm {\sqrt {m^{2}a^{2}+b^{2}}}}}\right),\,m\in \mathbb {R} .}$$

Here ${\displaystyle m}$ is the slope of the tangent at the corresponding ellipse point, ${\displaystyle {\vec {c}}_{+}}$ is the upper and ${\displaystyle {\vec {c}}_{-}}$ the lower half of the ellipse. The vertices${\displaystyle (\pm a,\,0)}$, having vertical tangents, are not covered by the representation.

The equation of the tangent at point ${\displaystyle {\vec {c}}_{\pm }(m)}$ has the form ${\displaystyle y=mx+n}$. The still unknown ${\displaystyle n}$ can be determined by inserting the coordinates of the corresponding ellipse point ${\displaystyle {\vec {c}}_{\pm }(m)}$:

$${\displaystyle y=mx\pm {\sqrt {m^{2}a^{2}+b^{2}}}\,.}$$

This description of the tangents of an ellipse is an essential tool for the determination of the orthoptic of an ellipse. The orthoptic article contains another proof, without differential calculus and trigonometric formulae.

Ellipse as an affine image of the unit circle

Another definition of an ellipse uses affine transformations:

Any ellipse is an affine image of the unit circle with equation ${\displaystyle x^{2}+y^{2}=1}$.

Parametric representation

An affine transformation of the Euclidean plane has the form ${\displaystyle {\vec {x}}\mapsto {\vec {f}}\!_{0}+A{\vec {x}}}$, where ${\displaystyle A}$ is a regular matrix (with non-zero determinant) 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 circle ${\displaystyle (\cos(t),\sin(t))}$, ${\displaystyle 0\leq t\leq 2\pi }$, is mapped onto the ellipse:

$${\displaystyle {\vec {x}}={\vec {p}}(t)={\vec {f}}\!_{0}+{\vec {f}}\!_{1}\cos t+{\vec {f}}\!_{2}\sin t\,.}$$

Here ${\displaystyle {\vec {f}}\!_{0}}$ is the center and ${\displaystyle {\vec {f}}\!_{1},\;{\vec {f}}\!_{2}}$ are the directions of two conjugate diameters, in general not perpendicular.

Vertices

The four vertices of the ellipse are ${\displaystyle {\vec {p}}(t_{0}),\;{\vec {p}}\left(t_{0}\pm {\tfrac {\pi }{2}}\right),\;{\vec {p}}\left(t_{0}+\pi \right)}$, for a parameter ${\displaystyle t=t_{0}}$ defined by:

$${\displaystyle \cot(2t_{0})={\frac {{\vec {f}}\!_{1}^{\,2}-{\vec {f}}\!_{2}^{\,2}}{2{\vec {f}}\!_{1}\cdot {\vec {f}}\!_{2}}}.}$$

(If ${\displaystyle {\vec {f}}\!_{1}\cdot {\vec {f}}\!_{2}=0}$, then ${\displaystyle t_{0}=0}$.) This is derived as follows. The tangent vector at point ${\displaystyle {\vec {p}}(t)}$ is:

$${\displaystyle {\vec {p}}\,'(t)=-{\vec {f}}\!_{1}\sin t+{\vec {f}}\!_{2}\cos t\ .}$$

At a vertex parameter ${\displaystyle t=t_{0}}$, the tangent is perpendicular to the major/minor axes, so:

$${\displaystyle 0={\vec {p}}'(t)\cdot \left({\vec {p}}(t)-{\vec {f}}\!_{0}\right)=\left(-{\vec {f}}\!_{1}\sin t+{\vec {f}}\!_{2}\cos t\right)\cdot \left({\vec {f}}\!_{1}\cos t+{\vec {f}}\!_{2}\sin t\right).}$$

Expanding and applying the identities ${\displaystyle \;\cos ^{2}t-\sin ^{2}t=\cos 2t,\ \ 2\sin t\cos t=\sin 2t\;}$ gives the equation for ${\displaystyle t=t_{0}\;.}$

Area

From Apollonios theorem (see below) one obtains:
The area of an ellipse ${\displaystyle \;{\vec {x}}={\vec {f}}_{0}+{\vec {f}}_{1}\cos t+{\vec {f}}_{2}\sin t\;}$ is

$${\displaystyle A=\pi \left|\det({\vec {f}}_{1},{\vec {f}}_{2})\right|.}$$

Semiaxes

With the abbreviations ${\displaystyle \;M={\vec {f}}_{1}^{2}+{\vec {f}}_{2}^{2},\ N=\left|\det({\vec {f}}_{1},{\vec {f}}_{2})\right|}$ the statements of Apollonios's theorem can be written as:

$${\displaystyle a^{2}+b^{2}=M,\quad ab=N\ .}$$

Solving this nonlinear system for ${\displaystyle a,b}$ yields the semiaxes:

$${\displaystyle {\begin{aligned}a&={\frac {1}{2}}({\sqrt {M+2N}}+{\sqrt {M-2N}})\\[1ex]b&={\frac {1}{2}}({\sqrt {M+2N}}-{\sqrt {M-2N}})\,.\end{aligned}}}$$

Implicit representation

Solving the parametric representation for ${\displaystyle \;\cos t,\sin t\;}$ by Cramer's rule and using ${\displaystyle \;\cos ^{2}t+\sin ^{2}t-1=0\;}$, one obtains the implicit representation

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

Conversely: If the equation

${\displaystyle x^{2}+2cxy+d^{2}y^{2}-e^{2}=0\ ,}$ with ${\displaystyle \;d^{2}-c^{2}>0\;,}$

of an ellipse centered at the origin is given, then the two vectors

$${\displaystyle {\vec {f}}_{1}={e \choose 0},\quad {\vec {f}}_{2}={\frac {e}{\sqrt {d^{2}-c^{2}}}}{-c \choose 1}}$$

point to two conjugate points and the tools developed above are applicable.

Example: For the ellipse with equation ${\displaystyle \;x^{2}+2xy+3y^{2}-1=0\;}$ the vectors are

$${\displaystyle {\vec {f}}_{1}={1 \choose 0},\quad {\vec {f}}_{2}={\frac {1}{\sqrt {2}}}{-1 \choose 1}.}$$

Whirls: nested, scaled and rotated ellipses. The spiral is not drawn: we see it as the locus of points where the ellipses are especially close to each other.

Rotated Standard ellipse

For ${\displaystyle {\vec {f}}_{0}={0 \choose 0},\;{\vec {f}}_{1}=a{\cos \theta \choose \sin \theta },\;{\vec {f}}_{2}=b{-\sin \theta \choose \;\cos \theta }}$ one obtains a parametric representation of the standard ellipse rotated by angle ${\displaystyle \theta }$:

$${\displaystyle {\begin{aligned}x&=x_{\theta }(t)=a\cos \theta \cos t-b\sin \theta \sin t\,,\\y&=y_{\theta }(t)=a\sin \theta \cos t+b\cos \theta \sin t\,.\end{aligned}}}$$

Ellipse in space

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

Polar coordinates centered at the center.

In polar coordinates, with the origin at the center of the ellipse and with the angular coordinate ${\displaystyle \theta }$ measured from the major axis, the ellipse's equation is[7]^: 75

$${\displaystyle r(\theta )={\frac {ab}{\sqrt {(b\cos \theta )^{2}+(a\sin \theta )^{2}}}}={\frac {b}{\sqrt {1-(e\cos \theta )^{2}}}}}$$

where ${\displaystyle e}$ is the eccentricity, not Euler's number

Polar coordinates centered at focus.

If instead we use polar coordinates with the origin at one focus, with the angular coordinate ${\displaystyle \theta =0}$ still measured from the major axis, the ellipse's equation is

$${\displaystyle r(\theta )={\frac {a(1-e^{2})}{1\pm e\cos \theta }}}$$

where the sign in the denominator is negative if the reference direction ${\displaystyle \theta =0}$ points towards the center (as illustrated on the right), and positive if that direction points away from the center.

In the slightly more general case of an ellipse with one focus at the origin and the other focus at angular coordinate ${\displaystyle \phi }$, the polar form is

$${\displaystyle r(\theta )={\frac {a(1-e^{2})}{1-e\cos(\theta -\phi )}}.}$$

The angle ${\displaystyle \theta }$ in these formulas is called the true anomaly of the point. The numerator of these formulas is the semi-latus rectum ${\displaystyle \ell =a(1-e^{2})}$.

Orthogonal diameters of a circle with a square of tangents, midpoints of parallel chords and an affine image, which is an ellipse with conjugate diameters, a parallelogram of tangents and midpoints of chords.

A circle has the following property:

The midpoints of parallel chords lie on a diameter.

An affine transformation preserves parallelism and midpoints of line segments, so this property is true for any ellipse. (Note that the parallel chords and the diameter are no longer orthogonal.)

Definition

Two diameters ${\displaystyle d_{1},\,d_{2}}$ of an ellipse are conjugate if the midpoints of chords parallel to ${\displaystyle d_{1}}$ lie on ${\displaystyle d_{2}\ .}$

From the diagram one finds:

Two diameters ${\displaystyle {\overline {P_{1}Q_{1}}},\,{\overline {P_{2}Q_{2}}}}$ of an ellipse are conjugate whenever the tangents at ${\displaystyle P_{1}}$ and ${\displaystyle Q_{1}}$ are parallel to ${\displaystyle {\overline {P_{2}Q_{2}}}}$.

Conjugate diameters in an ellipse generalize orthogonal diameters in a circle.

In the parametric equation for a general ellipse given above,

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

any pair of points ${\displaystyle {\vec {p}}(t),\ {\vec {p}}(t+\pi )}$ belong to a diameter, and the pair ${\displaystyle {\vec {p}}\left(t+{\tfrac {\pi }{2}}\right),\ {\vec {p}}\left(t-{\tfrac {\pi }{2}}\right)}$ belong to its conjugate diameter.

For the common parametric representation ${\displaystyle (a\cos t,b\sin t)}$ of the ellipse with equation ${\displaystyle {\tfrac {x^{2}}{a^{2}}}+{\tfrac {y^{2}}{b^{2}}}=1}$ one gets: The points

${\displaystyle (x_{1},y_{1})=(\pm a\cos t,\pm b\sin t)\quad }$ (signs: (+,+) or (−,−) )

${\displaystyle (x_{2},y_{2})=({\color {red}{\mp }}a\sin t,\pm b\cos t)\quad }$ (signs: (−,+) or (+,−) )

are conjugate and

${\displaystyle {\frac {x_{1}x_{2}}{a^{2}}}+{\frac {y_{1}y_{2}}{b^{2}}}=0\ .}$

In case of a circle the last equation collapses to ${\displaystyle x_{1}x_{2}+y_{1}y_{2}=0\ .}$

Theorem of Apollonios

For the alternative area formula

For an ellipse with semi-axes ${\displaystyle a,\,b}$ the following is true:[9][10]

Let ${\displaystyle c_{1}}$ and ${\displaystyle c_{2}}$ be halves of two conjugate diameters (see diagram) then

1. ${\displaystyle c_{1}^{2}+c_{2}^{2}=a^{2}+b^{2}}$.
2. The triangle ${\displaystyle O,P_{1},P_{2}}$ with sides ${\displaystyle c_{1},\,c_{2}}$ (see diagram) has the constant area ${\textstyle A_{\Delta }={\frac {1}{2}}ab}$, which can be expressed by ${\displaystyle A_{\Delta }={\tfrac {1}{2}}c_{2}d_{1}={\tfrac {1}{2}}c_{1}c_{2}\sin \alpha }$, too. ${\displaystyle d_{1}}$ is the altitude of point ${\displaystyle P_{1}}$ and ${\displaystyle \alpha }$ the angle between the half diameters. Hence the area of the ellipse (see section metric properties) can be written as ${\displaystyle A_{el}=\pi ab=\pi c_{2}d_{1}=\pi c_{1}c_{2}\sin \alpha }$.
3. The parallelogram of tangents adjacent to the given conjugate diameters has the ${\displaystyle {\text{Area}}_{12}=4ab\ .}$

Proof

Let the ellipse be in the canonical form with parametric equation

$${\displaystyle {\vec {p}}(t)=(a\cos t,\,b\sin t).}$$

The two points ${\textstyle {\vec {c}}_{1}={\vec {p}}(t),\ {\vec {c}}_{2}={\vec {p}}\left(t+{\frac {\pi }{2}}\right)}$ are on conjugate diameters (see previous section). From trigonometric formulae one obtains ${\displaystyle {\vec {c}}_{2}=(-a\sin t,\,b\cos t)^{\mathsf {T}}}$ and