BMLabs Mathematics

The section of the enveloping cone of the ellipsoid

\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1,

whose vertex is

P

, by the plane

z=0

is a rectangular hyperbola. The locus of

P

is to be found.

The section of the enveloping cone of the ellipsoid

\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1,

whose vertex is

P

, by the plane

z=0

is a rectangular hyperbola. The locus of

P

is to be found.

Let the coordinates of the vertex

P

(\alpha,\beta,\gamma)

. The equation of the enveloping cone of the ellipsoid is written as

\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1\right) \left(\frac{\alpha^2}{a^2}+\frac{\beta^2}{b^2}+\frac{\gamma^2}{c^2}-1\right) = \left(\frac{x\alpha}{a^2}+\frac{y\beta}{b^2}+\frac{z\gamma}{c^2}-1\right)^2.

Its section by the plane

z=0

is obtained by putting

z=0

in the above equation. The resulting curve lies in the

xy

-plane and is given to be a rectangular hyperbola. Hence, the sum of the coefficients of

x^2

and

y^2

must be zero. This condition yields

\begin{aligned} &\frac{1}{a^2}\left(\frac{\alpha^2}{a^2} +\frac{\beta^2}{b^2} +\frac{\gamma^2}{c^2}-1\right) -\frac{\alpha^2}{a^4} +\frac{1}{b^2}\left(\frac{\alpha^2}{a^2} +\frac{\beta^2}{b^2} +\frac{\gamma^2}{c^2}-1\right) -\frac{\beta^2}{b^4} =0. \end{aligned}

On simplification, this reduces to

\frac{1}{a^2}\left(\frac{\beta^2}{b^2}+\frac{\gamma^2}{c^2}\right) +\frac{1}{b^2}\left(\frac{\alpha^2}{a^2}+\frac{\gamma^2}{c^2}\right) =\frac{1}{a^2}+\frac{1}{b^2}.

Further simplification gives

c^2(\alpha^2+\beta^2)+(a^2+b^2)\gamma^2=c^2(a^2+b^2),

or,

\frac{\alpha^2+\beta^2}{a^2+b^2}+\frac{\gamma^2}{c^2}=1.

Hence, the locus of the vertex

(\alpha,\beta,\gamma)

\frac{x^2+y^2}{a^2+b^2}+\frac{z^2}{c^2}=1.

\quad

\blacksquare

The section of the enveloping cone of the ellipsoid

\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1

by the plane

z=0

is given to be a parabola. It is required to show that the locus of the vertex of the cone is the pair of planes

z=\pm c.

Let the vertex of the cone be

P(\alpha,\beta,\gamma)

. The equation of the enveloping cone corresponding to the ellipsoid is written as

\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1\right) \left(\frac{\alpha^2}{a^2}+\frac{\beta^2}{b^2}+\frac{\gamma^2}{c^2}-1\right) = \left(\frac{x\alpha}{a^2}+\frac{y\beta}{b^2}+\frac{z\gamma}{c^2}-1\right)^2.

Its section by the plane

z=0

is obtained as

\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}-1\right) \left(\frac{\alpha^2}{a^2}+\frac{\beta^2}{b^2}+\frac{\gamma^2}{c^2}-1\right) = \left(\frac{x\alpha}{a^2}+\frac{y\beta}{b^2}-1\right)^2.

Since the section is a parabola, the condition

(\text{coefficient of }2xy)^2 = (\text{coefficient of }x^2)\,(\text{coefficient of }y^2)

must be satisfied. Hence,

\begin{aligned} \left(-\frac{\alpha\beta}{a^2b^2}\right)^2 &= \left(\frac{\beta^2}{a^2b^2}+\frac{\gamma^2}{a^2c^2}-\frac{1}{a^2}\right) \left(\frac{\alpha^2}{a^2b^2}+\frac{\gamma^2}{b^2c^2}-\frac{1}{b^2}\right). \end{aligned}

On simplification, this leads to

\begin{aligned} \left(\frac{\alpha^2}{a^2}+\frac{\beta^2}{b^2}+\frac{\gamma^2}{c^2}-1\right) (\gamma^2-c^2)=0. \end{aligned}

Since the vertex lies outside the ellipsoid,

\frac{\alpha^2}{a^2}+\frac{\beta^2}{b^2}+\frac{\gamma^2}{c^2}\neq 1.

Therefore,

\gamma^2-c^2=0.

Hence, the locus of the vertex is

z^2=c^2 \quad \text{or} \quad z=\pm c.

\quad

\blacksquare

Find the locus of a luminous point if the ellipsoid

\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1

casts a circular shadow on the plane

z=0

Let the luminous point be

(\alpha,\beta,\gamma)

. This point serves as the vertex of the enveloping cone to the ellipsoid, whose truncated portion produces the shadow. The equation of the enveloping cone is

\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1\right) \left(\frac{\alpha^2}{a^2}+\frac{\beta^2}{b^2}+\frac{\gamma^2}{c^2}-1\right) = \left(\frac{x\alpha}{a^2}+\frac{y\beta}{b^2}+\frac{z\gamma}{c^2}-1\right)^2.

Its section by the plane

z=0

is given by

\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}-1\right) \left(\frac{\alpha^2}{a^2}+\frac{\beta^2}{b^2}+\frac{\gamma^2}{c^2}-1\right) = \left(\frac{x\alpha}{a^2}+\frac{y\beta}{b^2}-1\right)^2.

Since the section is a circle, the coefficient of

xy

must vanish and the coefficients of

x^2

and

y^2

must be equal. Hence,

\alpha\beta=0. \quad (1)

Also,

\frac{1}{a^2}\left(\frac{\beta^2}{b^2}+\frac{\gamma^2}{c^2}-1\right) = \frac{1}{b^2}\left(\frac{\alpha^2}{a^2}+\frac{\gamma^2}{c^2}-1\right). \quad (2)

\textit{Case I:}

\alpha=0

. From (2),

\begin{aligned} \frac{1}{a^2}\left(\frac{\beta^2}{b^2}+\frac{\gamma^2}{c^2}-1\right) &= \frac{1}{b^2}\left(\frac{\gamma^2}{c^2}-1\right), \end{aligned}

which reduces to

\frac{\beta^2}{b^2-a^2}+\frac{\gamma^2}{c^2}=1.

Hence, the locus is

\frac{y^2}{b^2-a^2}+\frac{z^2}{c^2}=1,\quad x=0.

\textit{Case II:}

\beta=0

. From (2),

\begin{aligned} \frac{1}{a^2}\left(\frac{\gamma^2}{c^2}-1\right) &= \frac{1}{b^2}\left(\frac{\alpha^2}{a^2}+\frac{\gamma^2}{c^2}-1\right), \end{aligned}

which gives

\frac{\alpha^2}{a^2-b^2}+\frac{\gamma^2}{c^2}=1.

Hence, the locus is

\frac{x^2}{a^2-b^2}+\frac{z^2}{c^2}=1,\quad y=0.

Thus, the required loci are obtained. \quad

\blacksquare

A variable plane is taken parallel to the plane

\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0

and it cuts the coordinate axes at the points

A,B,C

. It is required to prove that the circle

ABC

lies on the cone

\left(\frac{b}{c}+\frac{c}{b}\right)yz+\left(\frac{c}{a}+\frac{a}{c}\right)zx+\left(\frac{a}{b}+\frac{b}{a}\right)xy=0.

Without loss of generality, the equation of a plane parallel to the given plane may be written as

\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1,

where

a,b,c

are constants. Hence, the intercepts on the coordinate axes are obtained as

A(a,0,0), \quad B(0,b,0), \quad C(0,0,c).

The equations representing the circle

ABC

are taken as

\left. \begin{aligned} x^2+y^2+z^2-ax-by-cz &= 0,\\ \frac{x}{a}+\frac{y}{b}+\frac{z}{c} &= 1, \end{aligned} \right\} \quad (1)

where the first equation denotes the sphere passing through

O,A,B,C

, and the second denotes the plane containing the circle. Let a point

P(x,y,z)

be chosen on the required cone, and let the generator through

P

intersect the guiding circle

ABC

(\alpha,\beta,\gamma)

. Since

(\alpha,\beta,\gamma)

lies on the circle

ABC

, it satisfies

\alpha^2+\beta^2+\gamma^2-a\alpha-b\beta-c\gamma=0 \quad (2)

and

\frac{\alpha}{a}+\frac{\beta}{b}+\frac{\gamma}{c}=1. \quad (3)

The equations of the generator through

(\alpha,\beta,\gamma)

are expressed as

\frac{x}{\alpha}=\frac{y}{\beta}=\frac{z}{\gamma}=r,

or equivalently,

x=\alpha r,\quad y=\beta r,\quad z=\gamma r. \quad (4)

From equations (3) and (4), the parameter

r

is obtained as

\begin{aligned} \frac{x}{a}+\frac{y}{b}+\frac{z}{c} &= \frac{\alpha r}{a}+\frac{\beta r}{b}+\frac{\gamma r}{c} \\ &= r\left(\frac{\alpha}{a}+\frac{\beta}{b}+\frac{\gamma}{c}\right) \\ &= r. \end{aligned}

Hence,

r=\frac{x}{a}+\frac{y}{b}+\frac{z}{c}. \quad (5)

Again, using equations (2) and (4), it follows that

\begin{aligned} \frac{x^2+y^2+z^2}{r^2} &= \frac{ax+by+cz}{r}, \end{aligned}

which gives

x^2+y^2+z^2=r(ax+by+cz).

Substituting the value of

r

from (5), one obtains

\begin{aligned} x^2+y^2+z^2 &=\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)(ax+by+cz). \end{aligned}

On simplification, the equation reduces to

\left(\frac{b}{c}+\frac{c}{b}\right)yz +\left(\frac{c}{a}+\frac{a}{c}\right)zx +\left(\frac{a}{b}+\frac{b}{a}\right)xy=0.

Thus, the circle

ABC

is shown to lie on the above cone with vertex at the origin. \quad

\blacksquare

Solved Problems on Enveloping Cones and Locus of Vertex

Solved Problems on Enveloping Cones and Locus of Vertex