Vectors and Geometry Part 22

VECTORS AND GEOMETRY PART 22

PARABOLOIDS

OBJECTIVES

1 Familiarise the nature and shape of Paraboloids

2 Studies the concept of conjugate diametral planes

INTRODUCTION

In this session we discuss nature and shape of paraboloids especially the Elliptical Paraboloid . intersection of a line and a paraboloid ,equation to the normals ,condition of tangency etc are going to discuss .After the description of elliptical paraboloids ,cone through six normals ,conjugate diametral planes etc are discussing

Paraboloids

(A) The Elliptic paraboloid

The locus of the equation

… (1)

is called an elliptic paraboloid .now let’s discuss the nature and shape of this surface.

(i) No point bisects all chords through it and therefore there is no centre for the surface.

(ii) The coordinate planes, and bisect all chords perpendicular to them and are, therefore, its two planes of symmetry or two principal planes.

(iii) z cannot be negative (c being positive), and hence there is no part of the surface below the XOY planes, i.e., the surface is above the XOY plane.

(iv) The section by the plane is given by the equation

which represents an ellipse whose semi-axes are

and whose centre lies on the z-axis. It increases in size as k increases; there being no limit to the increase; for it is a point ellipse. The surface is therefore generated by a variable ellipse parallel to the XOY plane and is consequently called the elliptic paraboloid. Hence the surface is entirely on the positive side of the XOY plane, and extends to infinity.

The section of the surface by the plane is the parabola whose equations are

Thus the paraboloids also generated by a variable parabola in two different ways.

Note. When, the surface becomes a paraboloid of revolution, formed by revolving the parabola.

, about the z-axis.

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Intersection of a line with a paraboloid

To find the points of intersection of the line

, say… (2)

With the paraboloid

… (3)

The coordinates of any point on (2) are

If this point lies on (3), then

or…….(4)

This is quadratic in r and therefore gives two values of r, i.e., every line meets a paraboloid in two points.

If , then one root of this equation is infinite. Therefore, one point of intersection is at infinity, and the other point is at a finite distance.

Thus a line parallel to z-axis meets the paraboloid in one point at an infinite distance from and so meets it in one finite point only its distance is given by

Such a line is called a diameter of the paraboloid; and the point at finite distance is called the extremity of the diameter.

(a) Tangent lines and tangent plane

If the point lies on the surface, we have

which shows that one root of equation (4) of last article is zero. The other root is also zero, if in addition to it

touches the paraboloid at.

If we eliminate, between (3) and (4) we get the locus of all the tangent lines through, which is given by

i.e.,.

This is the equation of the tangent plane at to the paraboloid.

In particular, is the tangent plane at the origin and the z-axis is the normal .

(b) Condition of tangency

The condition that the plane

may touch the paraboloid

is,

and the point of contact, is

Corollary The plane

is a tangent plane to the paraboloid for all values of l, m, n.

(c) Locus of the point of intersection of the three mutually perpendicular tangent planes

Let the equations of the three mutually perpendicular tangent planes be

and

Adding these equations, we get

which is the required locus of the point of intersection of three mutually perpendicular tangent planes.

This is the plane at right angles to the z-axis, the axis of the paraboloid.

(d) Equations of normal

The tangent plane to the paraboloid

At any given point is given by

Therefore the equations of normal at are

(e) Normals from a given point

The equations normal at any point of the paraboloid

are

If it passes through the given point , say, then

, say

But lies on the paraboloid, therefore,

This is an equation of fifth degree in. it follows that there are five points on the paraboloid, the normals at which pass through the given point.

Thus, in general, five normals can be drawn through a given point to a paraboloid.

(f) Cubic curve through the feet of the normals

If the normal at to the paraboloid

Passes through a given point, we have as above.

Thus, the feet of the normals lie on the curve, its parametric equations is given by

… (5)

where is the parameter.

The point where this curve meets any given plane, say,

… (6)

are given by

This is a cubic in , giving three values of .

Therefore the plane (6) meets the curve (5) in three points, and hence it follows that the curve is a cubic curve.

(g) Cone through the five normals

If the normal at to the paraboloid passes through , then

Also the direction cosines of the normal at are proportional to .

If the line

… (7)

is a normal at , then

Hence

Therefore the locus of the normal (7) is

which is the equation of a cone.

Hence the five normals from to the paraboloid are generators of this cone.

(h) The polar plane of a point

Using the definition and notations , we have

i.e.,

i.e.,… (8)

Let the coordinates of R be , then

Hence (8) becomes

Therefore the locus of R is

which is the polar plane of with respect to the paraboloid

(i) Polar lines

The polar plane of any point

This plane for all values of r, passes through the line of intersection of the planes

and

These equations determine the polar of the given line.

(j) Section with a given centre

If be the middle point of the chord whose equations are

i.e,

Hence locus of all chords which are bisected at is the plane

This can be put in the form , where T and have their usual meaning.

(k) Locus of mid-point of a system of parallel chords

Let… (9)

Be a chord parallel to a fixed line

If is the middle point of the chord (9), then as before we have

It follows that the mid-point of all such chords lie in the plane

which is parallel to the z-axis, the axis of the paraboloid. The plane is called the diametral plane conjugate to the given direction.

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Conjugate diametral planes

We have already discussed in that the plane

… (10)

is the diameter plane of the line (Say OP)

with respect to the paraboloid

If OQ is another line whose equations are

and is parallel to the diametral plane (10) of OP, then we have

… (11)

The symmetry of this result shows that OP is parallel to the diametral plane of OQ, and therefore the diametral plane of OP and OQ are conjugate to each other.

Thus if and be two diametral planes, such that is parallel to the chords bisected by the plane, then is parallel to the chords bisected by. Two such planes are called conjugate diametral planes.

Equation (11) is the condition that the lines

;

and.

should be conjugate diameters of the conic

Hence any plane meets a pair conjugate diameter planes of a paraboloid in lines which are parallel to conjugate diameters of the conic in which the plane meets the surface.

Example 1 Two perpendicular tangent planes to the paraboloid

Intersect in straight line lying in the plane. Show that the line touches the parabola.

Solution

Equation of any line in the plane can be taken

Any plane through this line is

or… (12)

If this is a tangent plane to the paraboloid

then… (13)

This is quadratic in and therefore gives the values and thus two tangent planes through the line (9).

i.e. Also the two tangent planes are

and.

These planes are perpendicular if

i.e.

or… (14)

Now the required parabola is the envelope of line (10) subject to the condition (14). Eliminating p between (9) and (14) we have.

This is quadratic in the parameter.

Therefore the envelope of the line is given by

which is a parabola.

Example 2 Find the locus of the point of intersection of three mutually perpendicular tangent planes to the paraboloid

Solution

The plane touches the paraboloid, when or .

By putting the value of p, the plane reduces to

Since the three planes and hence their normals are mutually perpendiculars, we have

etc.

The locus of the point of intersection is obtained by eliminating from the equations of these planes.

For, we add these three planes obtained by putting and use the above relations. Thus we get

or.

The is the required plane.

Example 3 Find the condition that the Paraboloids

may have a common tangent plane.

Solution

Let the common tangent plane be

It will be a tangent plane to the given Paraboloids, if

Eliminating the unknown and 2np we get the required condition as

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Cone through the six normals

To show that the six normals from any point to the conicoid

lie on a cone of the second degree.

If the normal at to the given conicoid passes through , then

Also the direction cosines of the normal are proportional to .

Let the equations of a normal from be

… (15)

So that

… (16)

… (17)

and… (18)

Now multiplying (16) by , (17) by , and (18) by and adding, we get

This shows that the normal (15) is a generator of the cone

Hence the six normals from to the conicoids lie on a cone of second degree.

Example 4 If and are the feet of the six normals from a point to the ellipsoid.

and the plane is given by

then the plane is given by

Solution

Let the equation of the plane be

Then the feet of the six normals from any given point lie on the locus given by the equation.

… (19)

Let be a foot of the normals from any given point

, say

But lie on the given ellipsoid

This equation being a sixth degree equation in, gives six values of corresponding to the six feet of the normals.

Also putting the values of in the equation , we get

…..(20)

This is also sixth degree equation in , gives six values of corresponding to the six feet of the normals. Hence equation (19) and (20) are identical.

Therefore comparing coefficients of like terms in (19) and (20), we get

Substituting these values of the equation of plane becomes

or.

SUMMARY

Now its time to summarise the session in this session we discussed in detail the paraboloids ,its nature and shape .some of the properties like intersection of a line with a paraboloid ,Equation of normals,condition of tangency etc are discussed.cone through six normals , conjugate diametral planes are the other topics we discussed .before moving to next session let us try these questions

ASSIGNMENTS

1.Find the eqation of the normal at ( 4,3,5) on the paraboloid

2.Find the tangent plane to the paraboloid at the point

3. Find the equation of the tangent plane to the surface 2x2 -5y2=10z

FAQ

1.Discuss about a plane parallel to the axis of a paraboloid.

Ans; Any plane parallel to the axis of the paraboloid is easily seen, by the comparison, to be the diametral plane for the system of parallel chords with direction ratios

Any plane parallel to the axis of a paraboloid is therefore, a diametral plane.

QUIZ

1.The coordinates of the point of contact of the plane 8x-6y-z=5 and the paraboloid

(a)(8,9,5)(b)(1,0,0) (c)no point of contact(d)at the origin

2Equation of any diametral plane with respect to the paraboloid is

(a)alx+bmy+cn=0(b)alx-bmy+cn=0 (c)alx+bmy-cn=0 (d)alx-bmy-cn=0

Answers

1(a)(8,9,5)

2(c)alx+bmy-cn=0

GLOSSARY

ELLIPTIC PARABOLOID:The surface is therefore generated by a variable ellipse parallel to the XOY plane and is consequently called the elliptic paraboloid.

Diameter and extremity of the paraboloid:Thus a line parallel to z-axis meets the paraboloid in one point at an infinite distance from and so meets it in one finite point only its distance is given by

Such a line is called a diameter of the paraboloid; and the point at finite distance is called the extremity of the diameter

REFERENCE

1S.L.Loney The Elements of Coordinate Geometry ,Macmillian and company, London

2Gorakh Prasad and H.C.Gupta Text book of coordinate geometry, Pothisalapvt ltd Allahabad

3P.K.Mittal Analytic Geometry Vrinda Publication pvtLtd,Delhi.