Second-Order Surfaces. Center and Diametral Planes

Find the general equation of second-order surfaces cut by the plane $xOy$ in the curve $\alpha x^{2}+\beta xy+\gamma y^{2}+\delta x+\varepsilon y+\zeta = 0$.

Find the second-order surface on which lie the circles

Through the origin draw lines of asymptotic direction for the surface $x^{2}-2y^{2}-4xy-10xz-4yz-10 = 0$, parallel to the plane $3x+y-4z-3 = 0$.

Show that the equation $x^{2}+y^{2}+z^{2}+2xz+2y+1 = 0$ is satisfied only by the points of the line

Show that the equation

is satisfied by the coordinates of only one point $(-1,1,1)$.

Write a second-order equation that is satisfied only by the coordinates of the point $(2,1,1)$.

Set up a second-order equation that is satisfied only by the coordinates of the points of the line $2x+2 = 3y = 2z$.

Find the equation of the pair of planes $x+2y-z+1 = 0$ and $x-y+z+1 = 0$.

Find the lines along which the conical surface $z^{2}-2xy-4x-2y+2z-3 = 0$ intersects the plane $x+y+2z+5 = 0$.

Find the generators of the surface $x^{2}+y^{2}-z^{2} = 1$ passing through the point $(1,1,1)$.

Find the generator of the surface $x^{2}-y^{2} = 2z$ parallel to the plane $x+y+z = 0$.

Find the lines parallel to the generators of the two surfaces:

Through the points $(0,-2,2)$ and $(-1,0,0)$ draw planes intersecting the cone $x^{2}+y^{2} = z^{2}$ in parabolas.

Through the same points draw planes intersecting the same cone in ellipses.

Transform the equation $x^{2}+y^{2}+z^{2}+2xy-2yz+6xz+2x-6y-2z = 0$ by moving the origin to the center.

The same for the surface

The same for the surface $4xy+4xz-4y-4z-1 = 0$.

Find the locus of the centers of the surfaces $x^{2}+y^{2}-z^{2}+\lambda xz+\mu yz-2x-2y-2z = 0$.

From the center of the surface $x^{2}+4y^{2}-40z^{2}+4x-8y-16z-12 = 0$ draw lines of asymptotic direction.

Find the equation of the hyperboloid passing through the point $(2,1,1)$, given the asymptotic cone of this hyperboloid $x^{2}+y^{2}-z^{2} = 0$.

Prove that the locus of the centers of second-order surfaces passing through a given ellipse and through two points symmetric with respect to the plane of this ellipse is a second-order surface.

Denoting by $s_{\lambda \mu \nu }$ the left-hand side of the equation of the plane passing through the points $\lambda$ and $\mu$, prove that the equation

for any choice of the coefficients $A$, $B$, and $C$ represents a second-order surface passing through the six given points labeled $1,2,3,4,5$, and $6$.

The general equation of a second-order surface can be written in the form $Ax^{2}+By^{2}+Cz^{2}+Dxy+Exz+Fyz+Gx+Hy+Iz+K = 0$. By introducing a variable $t$, the left-hand side can be turned into a quadratic form in four variables

so that at $t = 1$ the equation of the surface is written as $\Phi (x,y,z,1) = 0$. At $t = 0$ the form $\Phi$ becomes the ternary quadratic form $f(x,y,z) = Ax^{2}+By^{2}+Cz^{2}+Dxy+Exz+Fyz$. The following identity holds (the Taylor formula for a quadratic form):

Using the Taylor formula (*), prove that when the origin is moved to the point $(\xi ,\eta ,\zeta )$ the equation of the surface $\Phi (x,y,z,1) = 0$ becomes:

If the determinant $\Delta_{3} = \begin{vmatrix} A & D & E \\ D & B & F \\ E & F & C \end{vmatrix}$ is nonzero, then it is possible to find $\xi$, $\eta$, and $\zeta$ for which $\dfrac {\partial \Phi }{\partial \xi } = 0$, $\dfrac {\partial \Phi }{\partial \eta } = 0$, $\dfrac {\partial \Phi }{\partial \zeta } = 0$. Prove that when the origin is moved to the point $(\xi ,\eta ,\zeta )$ the equation of the surface $\Phi (x,y,z,1) = 0$ becomes:

Here $\left.\dfrac {\partial \Phi }{\partial t}\right|_{t=1}$ is the partial derivative taken at $t = 1$. The indicated shift of origin is the transformation to the center.

If $(x,y,z)$ is the midpoint of a chord of the surface $\Phi (x,y,z,1) = 0$ (the notation $\Phi$ and $f$ is the same as in problem 577), and $(x+l\sigma ,y+m\sigma ,z+n\sigma )$ and $(x-l\sigma ,y-m\sigma ,z-n\sigma )$ are its endpoints, then the equality

holds. Prove this.

Prove that the midpoints of the chords of the surface $\Phi (x,y,z,1) = 0$ (the notation is the same as in problem 577) parallel to the vector $\overrightarrow {P}(l,m,n)$ lie on the plane

(The diametral plane conjugate to the direction of the vector $\overrightarrow {P}$.)

Prove that for the surface $f(x,y,z) \mp k = 0$ (the notation $f$ is the same as in problem 577) the equation of the diametral plane conjugate to the direction of the vector $\overrightarrow {P}(l,m,n)$ can be written in either of the two forms:

and

Prove that for the surface $\Phi (x,y,z,1) = 0$ (the notation is the same as in problem 577) the equation of the diametral plane conjugate to the direction of the vector $\overrightarrow {P}(l,m,n)$ can be written in the form:

Here $\dfrac {\partial f}{\partial l}$, $\dfrac {\partial f}{\partial m}$, and $\dfrac {\partial f}{\partial n}$ are the partial derivatives of $f(l,m,n)$, and $\left.\dfrac {\partial \Phi }{\partial t}\right|_{t=0}$ is the value of the quantity $\dfrac {\partial \Phi (l,m,n,t)}{\partial t}$ at $t = 0$.

Find the diametral plane of the surface $x^{2}+2y^{2}-z^{2}-2xy-2yz+2xz-4x-1 = 0$ conjugate to the direction of the vector $\overrightarrow {P}(1,-1,0)$.

Find the diametral plane of the surface $x^{2}+2y^{2}-z^{2}-2xy-2yz+2xz-4x-1 = 0$ parallel to the plane $x+y+z = 0$.

For the same surface find the diametral plane passing through the points $(0,0,0)$ and $(1,1,0)$.

Find the diametral plane of the surface $x^{2}+z^{2}-2x-1 = 0$ passing through the point $(2,1,0)$, as well as the direction conjugate to it.

Find the diametral plane of the surface $4x^{2}+y^{2}+z^{2}+4xy-4xz-2yz+x-y-1 = 0$ passing through the origin.

Find the equations of the diameter of the surface $4x^{2}+6y^{2}+4z^{2}+4xz+8y-4z+3 = 0$ perpendicular to the plane $xOz$.

The plane $x = 0$ is a diametral plane of the surface $xy = z$. Find the chords to which it is conjugate.

Find the diametral plane common to the surfaces $x^{2}+y+z = 0$ and $x^{2}+y^{2}-2x-2y-2z = 0$.

Find the equation of the diameter of the surface $x^{2}+2y^{2}-z^{2}-2xy-2yz+2xz-4x-1 = 0$ corresponding to the plane $x+y+z-6 = 0$.

Find the locus of the centers of the sections of the surface $x^{2}+2y^{2}+2z^{2}-2xy-2yz+2x-2z-1 = 0$ by planes parallel to the plane $x-y+z = 0$.

Given the diameter

of the surface $x^{2}+y^{2}+z^{2}-2yz-2x-y+1 = 0$, find the equations of the planes corresponding to it.