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For a second-degree equation $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$, the number ${\Delta }_2=B^2-4AC$ is the discriminant of a binary quadratic form associated with its quadratic part. For a nonempty, nondegenerate real conic section, a negative value gives an ellipse, zero a parabola, and a positive value a hyperbola. It is invariant under a rotation of orthonormal coordinates.^[ref]

A short explanation uses the symmetric matrixAn orthogonal matrix $R$ changes this matrix to $R^{\mathsf{T}}HR$, whose determinant equals $\det H$. The spectral theorem for real symmetric matrices then removes the mixed term. Equal-sign, opposite-sign and one-zero eigenvalues give the three types after the remaining linear terms and offsets are considered.

The qualification matters. The equations $x^2+y^2=0$, $y^2=0$ and $xy=0$ describe a point, a repeated straight line, and two crossing straight lines, respectively. Their quadratic discriminants have the signs expected of ellipse, parabola and hyperbola types, but all three are degenerate conics. Also, $x^2+y^2=-1$ has an empty set of real solutions despite its negative quadratic discriminant.

A directrix of a conic is the fixed line in its focus-directrix description: distance to the focus of a conic is a constant multiple of distance to this line. For a parabola the multiple is one. The directrix of a conic belongs to the geometry of the conic section, rather than being a tangent or an asymptote.

The parameter $e\ge 0$ in a focus-centred polar coordinates equation $r=\ell /(1-e\cos \theta )$ for a conic section. It distinguishes an ellipse ($e<1$), a parabola ($e=1$), and a hyperbola ($e>1$).

For a noncircular conic section, the eccentricity of a conic is the constant ratiowhere $F$ is a focus of a conic, $L$ the corresponding directrix of a conic, and $X$ a point on the curve. For an ellipse with semi-major axis $a$ and distance $c$ from the focal midpoint to either focus of a conic, $e=c/a=\sqrt{1-b^2/a^2}$. Its circle case has $e=0$; a finite directrix does not give that circle through the displayed ratio. The circle is the limit as $c\to 0$ and its directrices of a conic move arbitrarily far away.

The eccentricity of a conic is intrinsic to the cutting plane, and is not generally its slope in a chosen three-dimensional drawing. For the family $z=1+mx$ in conic section, with $0<m<1$, completing the square givesA displacement $dx$ along this plane has Euclidean norm $\sqrt{1+m^2}|dx|$. Thus the intrinsic semiaxis ratio satisfies $b^2/a^2=(1-m^2)/(1+m^2)$, givingFor $m=3/5$, this is $3/\sqrt{17}$, rather than $3/5$. This normalization separates the curve's geometry from a distorted projection of it onto the $xy$ plane.

In the focus-directrix description of a noncircular conic section, the focus is the fixed point $F$ for which every point $X$ of the curve satisfies $|X-F|=e\mathrm{dist}(X,L)$, where $L$ is a directrix of a conic and $e$ is the eccentricity of a conic. For $y^2=4ax$ with $a>0$, the focus of a conic is $(a,0)$, the directrix of a conic is $x=-a$, and $e=1$.

A hyperbola is a conic section with two unbounded branches. In suitable coordinates its equation is $x^2/a^2-y^2/b^2=1$ with $a,b>0$. The unit hyperbola is parametrized by $x=\cosh t$, $y=\sinh t$ on its right branch, relating it to the hyperbolic functions.

For example, $x^2/9-y^2/16=1$ has foci of a conic at $(\pm 5,0)$, because $5^2=3^2+4^2$. Its focal-distance difference has absolute value $6$, while its eccentricity of a conic is $5/3>1$. At the point $(3,0)$, the focal distances are $8$ and $2$, verifying that difference. The asymptotes are $y=\pm 4x/3$; a distant point approaches those directions while remaining on one of the two branches. The focal-distance difference, rather than a constant sum, distinguishes this hyperbola from an ellipse.

For the two centred hyperbolas $x^2-y^2=c$ and $4x^2-y^2=d$, with nonzero real $c,d$, eliminating their quadratic terms givesThus they meet in four real points if both quantities are positive, two if one is zero and the other positive, and none if either is negative. Both quantities cannot vanish when $c,d\ne 0$. The choices $(c,d)=(1,5),(1,4),(1,1)$ respectively realize these three counts. This reduction to a linear system in $x^2,y^2$ is useful for intersections of axis-aligned conic sections; the independent sign choices must be counted after solving that linear system.

Coordinates relative to an origin $\mathbf{o}$ and an orthonormal basis ${\mathbf{e}}_1,\dots ,{\mathbf{e}}_n$ preserve Euclidean distances: if $\mathbf{x}-\mathbf{y}={\sum }_i{d}_i{\mathbf{e}}_i$, then $\Vert \mathbf{x}-\mathbf{y}{\Vert }^2={\sum }_i{d}_i^2$. This matters when finding the eccentricity of a conic section in an oblique plane: arbitrary projected coordinates need not preserve the ratio of its semi-major axis to its semi-minor axis.

Write $k=GM$ and $r=|\mathbf{r}|$. The gravitational central force has zero torque, so the specific angular momentumis constant. For a nonradial orbit $h=|\mathbf{h}|>0$, motion lies in the plane normal to $\mathbf{h}$, and plane polar coordinates give $r^2\dot{\theta }=h$. With $u=1/r$ and primes denoting $\theta$ derivatives,The radial equation $\ddot{r}-r{\dot{\theta }}^2=-k/r^2$ is therefore the Binet equationIts general solution, after choosing the polar origin of angle, isThe origin is a focus of this conic section. To identify which conic section occurs, use the conserved specific energyA bound, nonradial orbit has $\epsilon <0$, hence $0\le e<1$ and is an ellipse. For example, putting $x=r\cos \theta$, $y=r\sin \theta$ in $r+ex=p$ givesan explicit ellipse equation. This is the intended low, bound Kepler orbit. The inverse-square equation alone does not imply an ellipse: an initially tangential speed larger than $\sqrt{2GM/r}$ gives $\epsilon >0$, $e>1$ and a hyperbolic Kepler orbit. Zero angular momentum instead gives radial motion. Those qualifications are necessary for the elliptic conclusion.

In a circular orbit, centripetal acceleration balances the gravitational acceleration:Here $\Omega$ is the positive angular speed; either sense of circulation has this period.

With the stated drag force, use the full angular momentum $\mathbf{L}=m\mathbf{r}\times \dot{\mathbf{r}}$. Gravitational torque is still zero, while the drag torque isTherefore $\boxed{\mathbf{L}(t)=\mathbf{L}(0)e^{-At/m}}$ for constant $A>0$. Its direction remains fixed, and its magnitude has exponential decay with time constant $m/A$; the orbit remains planar although its shape is no longer a fixed ellipse.

A nondegenerate hyperbola centred at the origin has a quadratic equationThere are no linear terms because the origin is its centre. Its asymptotes are the zero set of the homogeneous quadratic form $A{x}^2+2Bxy+C{y}^2$. Requiring the directions $y=x$ and $y=-x$ gives $A+2B+C=A-2B+C=0$, hence $B=0$ and $C=-A$. Since the conic section is nondegenerate, $A$ and $F$ are nonzero. Thus all such hyperbolas with prescribed asymptotes are exactlyFor $c>0$ these rectangular hyperbolas open along the $x$-axis, and for $c<0$ along the $y$-axis. Conversely, each displayed hyperbola has the required centre and asymptotes. Two distinct members have different values of $x^2-y^2$, so they cannot intersect.

For the second pair of asymptotes, the same argument gives $H_2:4x^2-y^2=d$, $d\ne 0$. The intersections of concentric hyperbolas therefore satisfyIf both right sides are positive, there are four independent sign choices. If precisely one is zero and the other is positive, there are two points. If either is negative there are none; both cannot vanish because $c\ne 0$. Each possibility occurs: take $(c,d)=(1,5)$ for four, $(1,4)$ for two, and $(1,1)$ for none. Hence

Set $X=x^2$ and $Y=y^2$ in the two conic sections. A real intersection requires $X,Y\ge 0$. Solving the resulting linear equations givesFor $a>0$ the denominator and $X$ are positive, so intersections exist exactly when $a\ge 1$. At $a=1$ they are $(\pm 1,0)$; for $a>1$ both choices of sign for each square root give intersections. Consequently

For $c\ne 0$, completing the square gives the conic sectionThe possible levels follow fromFor $0<c<1$, the contours are ellipses centred at $(0,1/c)$, shrinking to the maximum as $c\uparrow 1$. The zero contour is the parabola from part (c). For $-1<c<0$, the contours are vertical-opening hyperbolas. At $c=-1$ they degenerate into the intersecting straight lines, and for $-2<c<-1$ they are horizontal-opening hyperbolas. These descriptions specify the remaining contour families and their symmetry about the $y$ axis. The lower bound $-2$ is approached as $|x|\to \infty$ at fixed $y$ but is never attained.

For an attractive inverse-square force, write $V(r)=-\kappa /r$ with $\kappa >0$, and take nonzero conserved angular momentum $J$. The conic section orbit hasThe physically allowed cases are ellipse for $-m{\kappa }^2/(2J^2)\le E<0$, parabola for $E=0$, and hyperbola for $E>0$. The minimum allowed energy gives the circular special case $e=0$. Here $E$ is measured relative to infinity. Zero angular momentum instead gives radial collision or escape trajectories, interpreted as degenerate limits rather than regular closed ellipses.

For two bodies interacting through a potential energy that depends only on their separation, introduce the centre of mass coordinate and the two-particle relative coordinate:Their kinetic energy separates exactly:The centre of mass moves uniformly if there is no external force, while the internal two-body problem reduces to one effective particle of reduced mass $\mu$:The original trajectories follow from ${\mathbf{r}}_1=\mathbf{R}+(m_2/M)\mathbf{r}$ and ${\mathbf{r}}_2=\mathbf{R}-(m_1/M)\mathbf{r}$.

For Newtonian gravity, $U=-G{m}_1{m}_2/r$, so the relative acceleration is $\ddot{\mathbf{r}}=-GM\mathbf{r}/r^3$. Conserved relative angular momentum and energy then give the usual conic sections. For a bound orbit of relative semimajor axis $a$,The two bodies share this orbital period and orbit their common centre of mass with semimajor axes in inverse proportion to their masses. If $m_2\gg m_1$, then $\mu \simeq m_1$, recovering the fixed-heavy-centre approximation. Using $M$ as the internal inertial mass would confuse centre-of-mass translation with relative motion.

For a test mass moving around a fixed central mass $M$, gravity is central and conservative. Specific angular momentum $h=|\mathbf{r}\times \mathbf{v}|$ and specific orbital energy $ϵ=v^2/2-GM/r$ are constant. The motion is planar and sweeps equal areas in equal times, $dA/dt=h/2$.

For a nonradial orbit, the inverse-square force gives a conic section,Negative energy gives an ellipse, zero energy a parabola and positive energy a hyperbola. For an ellipse of semimajor axis $a$,A circular orbit is the $e=0$ case, with $v=\sqrt{GM/r}$. Escape from radius $r$ requires $v\ge \sqrt{2GM/r}$ in the absence of further forces. For two comparable masses, apply the relative two-body equation with $GM$ replaced by $G(M+m)$, rather than treating one body as fixed.

For a test particle outside a spherical central mass, the gravitational force is central, so angular momentum is conserved and the orbit is planar. With specific angular momentum $h=r^2\dot{\theta }$ and specific orbital energy $ϵ={\dot{r}}^2/2+h^2/(2r^2)-GM/r$, the radial effective potential combines attraction with the centrifugal barrier. Its nonradial solutions are conic sections:For nonzero $h$, negative energy gives an ellipse with $0\le e<1$, zero energy a parabolic Kepler orbit and positive energy a hyperbolic Kepler orbit. The circle is the special case $e=0$. Exactly radial motion is a separate zero-angular-momentum case and may hit the central body; a mathematical point-mass trajectory is not automatically a physically noncolliding orbit.

Conservation of $h$ gives constant areal velocity $h/2$, so equal areas are swept in equal times. For a bound ellipse of semimajor axis $a$, $ϵ=-GM/(2a)$ and $T^2=4{\pi }^2{a}^3/(GM)$. A circular orbit has speed $\sqrt{GM/r}$, while the minimum local escape speed is $\sqrt{2GM/r}$, obtained by setting the total energy to zero. These are the test-mass two-body results: a comparable second mass requires the relative-coordinate gravitational parameter $G(M+m)$, and nonspherical structure or drag can perturb the conic and its conserved quantities.

In a fixed point-mass Newtonian gravitational field, a test mass $m$ experiencesBecause this is a central force, its torque about the attracting centre vanishes and angular momentum is conserved. The orbit lies in a fixed plane and sweeps out equal areas in equal times. Since the force is also conservative, the specific orbital energy is constant:The last two terms form an effective potential. They show how the angular-momentum barrier competes with attraction and determines the turning radii. For nonradial motion the orbit is a conic section:Negative energy gives an ellipse, with $e<1$; zero energy gives a parabola; positive energy gives a hyperbola. The circular limit has $e=0$, $v_c=\sqrt{GM/r}$ and $\epsilon =-GM/(2r)$. For a bound ellipse of semimajor axis $a$, $\epsilon =-GM/(2a)$ and the orbital period is $T=2\pi \sqrt{a^3/(GM)}$. The local escape speed is $\sqrt{2GM/r}$. These are the point-mass, test-particle limits; extended mass distributions or noncentral perturbations change the orbit laws.

For a light test particle of mass $m$ outside a spherical source of mass $M$, Newton's law of universal gravitation gives $\mathbf{F}=-GMm\hat{\mathbf{r}}/r^2$. It is a central force, so its torque about the source vanishes and angular momentum $\mathbf{L}=m\mathbf{r}\times \mathbf{v}$ is conserved. The motion therefore lies in a fixed plane when $L\ne 0$. Its gravitational potential energy is $-GMm/r$, and conservation of mechanical energy givesHere $ϵ$ is specific orbital energy and $h$ is specific angular momentum. A Kepler orbit has polar equationThe angle is measured from periapsis. The conic section is an ellipse for $ϵ<0$ and $0\le e<1$, a parabola for $ϵ=0$ and $e=1$, and a hyperbola for $ϵ>0$ and $e>1$. The circle is the $e=0$ elliptical case. The special case $h=0$ is radial motion and must be handled as a degenerate limit. Bound ellipses have the source at one focus, which is Kepler's first law.

For a bound orbit of semi-major axis $a$, $ϵ=-GM/(2a)$. Combining this with the energy equation gives the vis-viva equation:The particle therefore moves faster near periapsis and slower near apoapsis. Conservation of angular momentum also gives constant areal velocity, $dA/dt=h/2$, expressing Kepler's second law. Using the ellipse's area and $h^2=GMa(1-e^2)$ gives the orbital periodwhich is Kepler's third law.

A circular orbit follows from centripetal acceleration $v_c^2/r=GM/r^2$, giving $v_c=\sqrt{GM/r}$ and $T=2\pi \sqrt{r^3/(GM)}$. The minimum escape velocity from radius $r$ follows from the marginally unbound condition $ϵ=0$:These ideal orbits neglect drag force, other gravitating bodies and changes in the central source. For two comparable masses, the relative two-body problem uses $G(M+m)$ instead of $GM$. An orbiting object is continuously in free fall; the inward gravitational force changes its direction even when its speed is constant. These inverse-square results assume the physical variation of gravity with radius, rather than extending a near-surface constant-$g$ approximation to an entire orbit.

A conic section with a focus of a conic at the origin has the polar coordinates equation $r=\ell /(1+e\cos \theta )$, where $\ell >0$ is the semilatus rectum and $e\ge 0$ the eccentricity. The directrix of a conic definition $r=e(d-r\cos \theta )$ gives the formula with $\ell =ed$ when $e>0$; $e=0$ is the circle limit. The cases $0<e<1$, $e=1$, and $e>1$ give an ellipse, parabola, and hyperbola. The physical radius requires a positive denominator. For an attractive inverse-square central force, the Laplace-Runge-Lenz vector gives $\ell =m{h}^2/k$ and $e=A/(mk)$.

The distance from a focus to a conic section along a line perpendicular to its symmetry axis. In focus-centred polar coordinates, an ellipse may be written $r(1-e\cos \theta )=\ell$, where $e$ is its eccentricity of a conic.

The distance from a focus of a conic section to the curve measured perpendicular to its principal axis. In the focus-based polar coordinates equation $r=p/(1+e\cos \varphi )$ it is the length $p$, where $e$ is the eccentricity. For an ellipse, $p=a(1-e^2)=b^2/a$. For a Newtonian Kepler orbit, $p=j^2/(GM)$ with $j$ the specific angular momentum.

The semi-major and semi-minor axes are the larger and smaller half-axis lengths of an ellipse. In principal-axis coordinates the ellipse has equation $x^2/a^2+y^2/b^2=1$ with $a\ge b>0$. The same terminology is used for a hyperbola $x^2/a^2-y^2/b^2=1$, where $a$ is half its transverse axis and $b$ half its conjugate axis. Orbital semiaxes apply these geometric quantities to conic sections.