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An elliptic cone is a real quadric surface consisting of the two nappes $u=\pm \sqrt{(b{v}^2+c{w}^2)/a}$, joined at the vertex. Its intersections with planes $u=u_0\ne 0$ are ellipses. It is the zero level of an indefinite quadratic form of signature $(1,2)$. Changing the level to a positive value produces a two-sheeted hyperboloid; changing it to a negative value produces a one-sheet hyperboloid, when the positive coefficient is assigned to the axial coordinate $u$.

Set $x=a\cosh \rho \cos u\cos v$, $y=a\cosh \rho \cos u\sin v$, $z=a\sinh \rho \sin u$, with $a>0$, $\rho \ge 0$ and $|u|\le \pi /2$. Differentiating the position vector gives pairwise orthogonal coordinate tangents and the displayed scale factors of orthogonal coordinates. Thus the positive volume density is $a^3\cosh \rho \cos u({\sinh }^2\rho +{\sin }^{2}u)$. The ordered Jacobian determinant is negative for the order $(\rho ,u,v)$; volume uses its absolute value. Positive constant $\rho$ surfaces are oblate spheroids. For $0<|u|<\pi /2$, constant $u$ gives the half of the one-sheet hyperboloid $(x^2+y^2)/(a^2{\cos }^{2}u)-z^2/(a^2{\sin }^{2}u)=1$ with the sign of $z$ fixed by $u$. At $\rho =0$ the coordinates cover the equatorial disk, while $u=0$ covers the exterior equatorial plane.

The displayed surfaces, with positive semiaxes, are hyperboloids. The plus sign gives a one-sheet hyperboloid, while the minus sign gives a two-sheeted hyperboloid.

Split the mixed quadratic terms evenly across a real symmetric matrix:Its characteristic polynomial isso $c=2$. Choose the orthonormal eigenvectorsThe orthogonal transformation $\mathbf{r}=P\mathbf{R}$ givesIndeed $P^{T}P=I$ and $P^{T}AP=\mathrm{diag}(-1,-5,4)$; preserving $Q_2$ is precisely the length-preserving property of an orthogonal matrix.

The requested plane has normal $(2,1,-2)$. An orthonormal basis within it isWrite a point of that plane as $\mathbf{r}=u{e}_1+w{e}_2$. Restricting the quadratic form gives $Q_1=-u^2-(13/9)w^2$, so its section with $Q_1=-5$ isIt is an ellipse centred at the origin, with semi-axes $\sqrt{5}$ along $e_1$ and $\sqrt{45/13}$ along $e_2$. Normalizing the in-plane basis is essential: coefficients in nonorthonormal coordinates would not directly give physical lengths.

The complete surface is the one-sheet hyperboloidIt is connected, extends indefinitely in both directions along its axis $u_3$, and has an elliptical waist at $Z=0$ with semi-axes $\sqrt{5}$ and $1$. At larger $|Z|$ its transverse sections widen; its asymptotic cone is $4Z^2=X^2+5Y^2$. The specified plane is oblique to that axis, rather than the waist plane.

For $A^{\prime }$, the centre and constant remain $\mathbf{a}=(-1,0,0)$ and $C=3$. Its eigenvalues are $3,1,-1$, so rotated coordinates put the surface in the form $3u^2+w^2-y^2=3$. This is a one-sheet hyperboloid. Since $3\Vert \mathbf{x}{\Vert }^2\ge {\mathbf{x}}^{\mathsf{T}}{A}^{\prime }\mathbf{x}=3$, the nearest centre-to-surface distance is at least one, attained in the eigen-direction with eigenvalue three. The coordinate $y$ is unbounded, because arbitrarily large $|y|$ can be balanced by $u$ or $w$. Thus the nearest distance is one, and there is no maximum value of $y$.

For the printed nonsymmetric $A^{\prime \prime }$, use the symmetric part of a matrix:The skew-symmetric part contributes zero to this scalar. The same centre solves $2S\mathbf{a}+\mathbf{b}=0$, and again $C=3$. Sylvester's positive principal minors are $2,2,5/2$, so this surface is an ellipsoid.

Let ${\lambda }_{\ast }$ be the largest eigenvalue of $S$. Its characteristic equation is $4{\lambda }^3-20{\lambda }^2+27\lambda -10=0$, and the Rayleigh quotient givesFor the maximum $y$, the support function of a centred ellipsoid gives the bound below. Equivalently, apply the Cauchy-Schwarz inequality in the $S$-weighted inner product: $y^2\le 3{\mathbf{e}}_y^{\mathsf{T}}{S}^{-1}{\mathbf{e}}_y$. The inverse entry is ${\mathbf{e}}_y^{\mathsf{T}}{S}^{-1}{\mathbf{e}}_y=6/5$, and equality is attained for a centred vector proportional to $S^{-1}{\mathbf{e}}_y$. Thus

Using the eigenvalues of the raw nonsymmetric $A^{\prime \prime }$ would be incorrect: it is its symmetric part that determines the surface and its Euclidean geometry.

In the standard spherical coordinate system, $\theta$ is measured from the positive $z$ axis. The radial unit vector in spherical coordinates isThe coordinate forms of the elliptic paraboloid are given in the labelled parts below.

For the other surface, $r^2\cos 2\theta =r^2({\cos }^2\theta -{\sin }^2\theta )$ converts toFor $a>0$, this is a two-sheeted hyperboloid, with two disconnected sheets starting at $z=\pm \sqrt{a}$ and opening along the $z$ axis. For $a<0$, it is a one-sheet hyperboloid:Its waist is the circle $z=0$, $x^2+y^2=|a|$. Both are surfaces of revolution about the $z$ axis and approach the cone $z^2=x^2+y^2$ far from the origin.

The figure shows $a=1$ and $a=-1$; scaling all coordinates by $\sqrt{|a|}$ gives the other magnitudes.

Represent the quadratic form by the symmetric matrixThe $xy$ block has characteristic polynomial $(2-\mu )(5-\mu )-4=(\mu -1)(\mu -6)$. Corresponding orthonormal eigenvectors, together with the $z$ direction, areThe eigenvalues are $1,6,-1$. To eliminate the linear term, solve $H\tilde{O}=-(0,3\sqrt{5},0{)}^T$, obtainingIndeed the first equation gives $x=y$, and the second then gives $3y=-3\sqrt{5}$. Define the new coordinates by $X=\tilde{O}+\tilde{x}{\tilde{e}}_1+\tilde{y}{\tilde{e}}_2+\tilde{z}{\tilde{e}}_3$. The original polynomial takes the value $-15$ at the center, so its translated and orthogonally diagonalized form isThusThe surface is an elliptic one-sheet hyperboloid, centered at $\tilde{O}$, with axis parallel to ${\tilde{e}}_3$. Its waist at $\tilde{z}=0$ is an ellipse with semiaxes $\sqrt{15}$ and $\sqrt{5/2}$. Every constant-$\tilde{z}$ section is a nonempty ellipse, so the two signs of $\tilde{z}$ belong to the same connected sheet. This is an application of principal-axis reduction of a quadric.

Stokes theorem equates the circulation around the induced boundary orientation of an oriented surface to the normal flux of curl:The usual version requires a piecewise smooth oriented surface and a vector field continuously differentiable on a neighbourhood of it. We use the upward orientation wherever the surface is a graph; each outer circle is counterclockwise viewed from above, and an inner circle is clockwise.

Write $\rho =(x^2+y^2{)}^{1/2}$. The surface of revolution has $z=\sqrt{{\rho }^2+1-\lambda }$ with $0\le z\le 1$. For $0<\lambda <1$, it is a disk-shaped cap of a two-sheeted hyperboloid, running from $\rho =0,z=\sqrt{1-\lambda }$ to $\rho =\sqrt{\lambda },z=1$. For $\lambda =1$, it is a right circular cone, $z=\rho$, including its apex. For $\lambda >1$, it is an annular strip of a one-sheet hyperboloid, between the circles $\rho =\sqrt{\lambda -1},z=0$ and $\rho =\sqrt{\lambda },z=1$.

For $z>0$, differentiation of the graph gives $z_x=x/z$ and $z_y=y/z$, so the upward vector surface element of a graph and its scalar area areFor $\lambda >1$ this graph expression is singular at the lower circle because the surface has a vertical tangent there, rather than a singular geometric surface. A regular parametrization isIt proves that the graph surface integral is interpreted by its integrable endpoint limit. The same parametrization works away from the pole or apex in the other regimes.

Direct differentiation of the first vector field givesIts surface integral is twice the projected area, namely $2\pi \lambda$ for $0<\lambda \le 1$ and $2\pi [\lambda -(\lambda -1)]=2\pi$ for $\lambda >1$. On an oriented circle of radius $R$, $\mathbf{F}\cdot d\mathbf{x}=R^{2}d\varphi$. The outer circulation is $2\pi \lambda$; an inner circle, when present, contributes $-2\pi (\lambda -1)$. ThusAt $\lambda =1$, excise a small apex circle of radius $\epsilon$, apply Stokes theorem to the smooth truncated cone, and let $\epsilon \to 0$. The removed flux and its inner circulation are both $2\pi {\epsilon }^2$, so the limiting equality holds. This avoids silently applying a smooth-surface theorem at the cone's nonsmooth apex.

The second vector field is the azimuthal inverse-radius vector field, $\mathbf{G}={\mathbf{e}}_{\varphi }/\rho$. Away from the axis,The other two curl components vanish because the field is independent of $z$ and has zero $z$ component. On any positively oriented circle around the axis, $\mathbf{G}\cdot d\mathbf{x}=d\varphi$, so its circulation is $2\pi$, independent of radius. ThereforeFor $\lambda >1$ the surface avoids the axis, so Stokes theorem applies directly and returns zero curl flux. For $0<\lambda <1$ the disk cap meets the axis, and for $\lambda =1$ the apex lies on it: $\mathbf{G}$ is undefined there, so the neighbourhood hypothesis of Stokes theorem fails. Removing a small circle gives cancelling inner and outer circulations; the inner circulation remains $-2\pi$ as its radius tends to zero. Hence it cannot be discarded as it could for $\mathbf{F}$. A nonzero circulation here is compatible with zero curl away from the excluded axis.

On the part with $1\le z\le 2$, writeThis is the upper portion of a hyperboloid of one sheet. Choosing the outward orientation, whose normal points radially away from the $z$-axis, the vector area element isIts radial component is positive and its vertical component is negative. Reversing the normal reverses all the fluxes below.

Forthe curl isOn the surface, $z^2={\rho }^2-1$, and thereforeDirect integration gives

To verify this with the Stokes theorem, note that on a circle of fixed $\rho ,z$,The induced boundary orientation runs in the negative $\varphi$ direction on the top circle and the positive $\varphi$ direction on the bottom circle. Hence

For $S^{\prime }$, Stokes' theorem again reduces the flux to its two boundary circles. The outward orientation gives positive $\varphi$ direction at $z=-1$, where ${\rho }^2=2$, and negative $\varphi$ direction at $z=\sqrt{2}$, where ${\rho }^2=3$. Thus

The planemeets the one-sheet hyperboloid in the two disjoint curvesThey are the two opposite meridians of the surface of revolution. Parametrize the generating profile by arc length. Its acceleration lies in the meridian plane and is orthogonal to the profile tangent; it is also orthogonal to the azimuthal tangent. It is therefore normal to the surface, so both meridians are geodesics. This is the disjoint case in plane-section geodesics of the unit one-sheet hyperboloid.

An allowable parametrisation is a smooth homeomorphism from an open subset of ${\mathbb{R}}^2$ onto an open subset of $S$, with derivative of rank two. Away from the axis, the rotation orbit has nonzero tangent. A transverse curve supplied by the submanifold theorem, followed by the rotation action, giveswith $|v|<\pi$ and $(f^{\prime },g^{\prime })\ne (0,0)$.

For a ruled parametrisation, regularity is exactly

Rotate and translate along the axis so the specified ruling isHere $s\ne 0$ because the line is not parallel to the axis, and $c\ne 0$: if $c=0$, rotating the horizontal tangent line makes the ruled parametrisation singular at its closest point. Rotating $L$ givesa one-sheet hyperboloid. Rotation invariance puts this whole surface in $\Sigma$. Connectedness and the fact that a complete embedded hyperboloid cannot be a proper subset of another connected embedded surface force equality. Rescaling radial and axial coordinates gives a diffeomorphism with $x^2+y^2=1+z^2$.

Choose the rotated Cartesian coordinatesTheir inverse is $x=(X-Y)/\sqrt{2}$, $y=(X+Y)/\sqrt{2}$. Consequently $x^2+y^2=X^2+Y^2$ and $xy=(X^2-Y^2)/2$, giving the canonical form of a quadricThe orthogonal diagonalization identifies the eigenvalues $1+a/2,1-a/2,1$ of the symmetric matrix of the quadratic form.

For $|a|<2$, all three coefficients are positive: the surface is an ellipsoid with semiaxes $(1+a/2{)}^{-1/2},(1-a/2{)}^{-1/2},1$. At $a=0$ it is a sphere.

For $a=2$, the equation is $2X^2+Z^2=1$, an elliptic cylinder parallel to the $Y$ axis. For $a=-2$, it is $2Y^2+Z^2=1$, an elliptic cylinder parallel to the $X$ axis. The absent coordinate is unrestricted; its disappearance does not turn the surface into an ellipse.

For $a>2$, the negative coefficient is that of $Y^2$: this is a one-sheet hyperboloid with axis $Y$. For $a<-2$, the negative coefficient is that of $X^2$: it is a one-sheet hyperboloid with axis $X$. This gives the signature transition from an ellipsoid to an elliptic cylinder and a one-sheet hyperboloid, including both transition values.

Let ${\lambda }_1={\lambda }_2=\lambda >0$ and ${\lambda }_3=-\gamma <0$. ThenThis is a one-sheet hyperboloid of revolution around the $x_3^{\prime }$ axis. Its circular waist has radius $\sqrt{k/\lambda }$ at $x_3^{\prime }=0$, and its radius grows with $|x_3^{\prime }|$. In the original coordinates the axis is the eigenvector direction belonging to the negative eigenvalue.

The matrix is $M=2I-e{e}^T$, where $e=(1,1,1{)}^T$. Along $\hat{e}=e/\sqrt{3}$, its eigenvalue is $2-3=-1$. Every vector perpendicular to $e$ has eigenvalue two. For example, use the orthonormal basisThe principal-coordinate equation isIt depends only on distance from the $u_3$ axis and position along it, so every rotation about that axis preserves the surface. ThereforeThis is a one-sheet hyperboloid with waist radius $\sqrt{3/2}$. The repeated positive eigenvalue makes the two transverse principal axes nonunique.

Use the principal coordinatesPart (c) puts the equation in the canonical formIf $c>2$, all three coefficients are positive: the surface is an ellipsoid with semiaxes $1/\sqrt{6}$, $1/\sqrt{3}$ and $1/\sqrt{2c-4}$ along the three eigenvectors listed in part (c).

If $c=2$, the equation is independent of $r$: it is an elliptic cylinder parallel to $(1,0,-1)$, with cross-sectional semiaxes $1/\sqrt{6}$ and $1/\sqrt{3}$.

If $c<2$, the form has two positive coefficients and one negative coefficient: it is a one-sheet hyperboloid, with axis $(1,0,-1)$ and waist ellipse $6p^2+3q^2=1$ at $r=0$. Its axial scale is $1/\sqrt{4-2c}$.

These are the three cases of the signature transition from an ellipsoid to an elliptic cylinder and a one-sheet hyperboloid. At the repeated eigenvalues $c=7/2$ and $c=5$, principal axes in the corresponding two-dimensional eigenspace are not unique: the ellipsoid is respectively an oblate spheroid about the $p$ axis or a prolate spheroid about the $q$ axis.

Put $s^2=x^2+y^2$. For $\rho >0$, elimination of $u$ givesThese are oblate spheroids with equatorial radius $\cosh \rho$ and polar radius $\sinh \rho$, flattened along the $z$ axis. At $\rho =0$ the limiting surface is the disk $z=0$, $s\le 1$.

For $0<|u|<\pi /2$, elimination of $\rho$ instead givesThese are one-sheet hyperboloids of revolution, restricted to their upper half for $u>0$ and lower half for $u<0$. The restriction matters because $\rho \ge 0$. At $u=0$, the surface degenerates to the exterior equatorial plane $z=0$, $s\ge 1$; at $u=\pm \pi /2$ it degenerates to the corresponding half of the $z$ axis.

Finally, fixed $v$ specifies a vertical half-plane through the $z$ axis, with horizontal direction $(\cos v,\sin v)$. The two values $v$ and $v+\pi$ together give a full plane. These surfaces form the mutually orthogonal families of the hyperbolic parametrization of oblate spheroidal coordinates.

In the orthonormal basis of part (b), the three quadric surfaces have the following forms. For $\alpha =1$,a two-sheeted hyperboloid with axis $u$, vertices $(\pm 1/\sqrt{6},0,0)$, and one sheet on each side of the plane $u=0$. For $\alpha =0$,an elliptic cone with vertex at the origin and the same axis. For $\alpha =-1$,a one-sheet hyperboloid with axis $u$ and a waist ellipse at $u=0$, whose semiaxes are $1/\sqrt{3}$ along $v$ and $1/2$ along $w$. The sections at fixed $u$ enlarge away from the waist.

Write a real quadratic polynomial as $x^{T}Hx+2q^{T}x+c$, with $H$ symmetric. If $H$ is invertible, translation by $x_0=-H^{-1}q$ removes the linear term. An orthonormal eigenbasis then makes the remaining quadratic form diagonal. Eigenvalue signs and the translated constant determine the real geometry, such as an ellipsoid or a one-sheet hyperboloid. A singular $H$ requires separate treatment of linear terms in its null directions; a center need not exist.

After an orthogonal diagonalization, the sign of the third coefficient classifies the displayed quadric surface. If $C>0$, it is an ellipsoid with semiaxes $A^{-1/2},B^{-1/2},C^{-1/2}$. If $C=0$, it is an elliptic cylinder parallel to the $w$ axis, with elliptical transverse section. If $C<0$, it is a one-sheet hyperboloid with axis $w$ and waist ellipse $A{u}^2+B{v}^2=1$. The continuous limit $C\to 0$ sends the associated axial length scale to infinity. Repeated positive eigenvalues make the corresponding principal-axis choices nonunique within their eigenspace, without changing the surface.

This regular surface has two connected components, $t=\pm \sqrt{1+x^2+y^2}$. The symmetric-matrix model of the two-sheeted hyperboloid realizes them as the positive-definite and negative-definite symmetric determinant-one $2\times 2$ matrices. Each component is diffeomorphic to ${\mathbb{R}}^2$; this differs from the connected one-sheet hyperboloid.