Substituting .. this value of x in equation (1) we find

To find the polar equation to the hyperbola, the centre being the pole.

CM = x

PCA=0

MP=y.

From the right angled triangle CPM, we have

222+y2, substituting

for y' its value derived from the equation to curve

P

A

AM

To find the equation to the hyperbola referred to its asymptotes as axes of co-ordinates.

It has been already shown that the asymptotes of the hyperbola are diameters of the curve, and that they are the diagonals of the rectangle ABA'B', whose sides are equal the major and minor axes of the curve.

Now C being the origin of co-ordinates, the equation to the straight line CZ is of the form

y = ax

where a is the tangent of ZCX.

C

And in like manner the equation to the asymptote zCz' is

(M)

(N)

prefer however, in this case, to perform the operation independently. Let CZ be assumed as the new axis of y's, and CZ as the new axis of r's, let the angle ZCX which they make with CX be called .

Take any point P in the curve, draw PM perpendicular to CX; Pm parallel to CZ; PR parallel to CZ'; RN perpendicular to CX; PQ parallel to CX. Assume CM= *, MP =3

Cm = X, mPY, angle ZCX = 4

Now,

Substituting these values of x and y in equation (O), it becomes

a2 (Y -X)2 sin.2 · b2 (Y + X)2 cos.2 = -a2b2.

Now we have seen above, that

tan. ZCX or =

a2 + b2
and cos 2p those values thus derivea,

Substituting ... for sin.2
a262
a2 + b2 (Y — X)2 —

a262

a2 + b2

Or, changing the large letters, which we no longer require for distinction,

a2 + b2 ry =

which is the equation to the hyperbola referred to its asymptotes.

EXERCISES IN ANALYTICAL GEOMETRY.

(1.) Construct the equations 5y-3x-2=0, 8y — 6x+5=0, y + 3x = 6, y2 —5x2 = 0, and y2 — 7y + 12 = 0; the axes of co-ordinates being rectangular.

(2.) Find the equation to the straight line which passes through the two points (2, 3) and (4, 5). Ans. y=x+ 1.

(3.) Describe the circle whose equation is y2 + x2 + 4y — 4x = 8.

(4.) Find the co-ordinates x'y' of the centre, and R the radius of the circle whose equation is

y2 + x2 - 6y+8x-11=0.

Ans. y'3, x' =— 4 and R6.

(5.) Prove that the perpendiculars drawn from the angles to the opposite sides of a triangle pass through the same point.

(6.) Prove that the straight lines drawn from the angles of a triangle, to bisect the opposite sides, pass through the same point.

(7.) Given the base = a, and the sum of the squares of the sides mine the locus of the vertex of the triangle.

(8.) Given the base of a triangle the locus of the vertex.

=

s2, to deter

a, and the ratio of the sides mn, to find

(9.) Given the base and the vertical angle, to determine the locus of the vertex of the triangle.

(10.) From a given point A, either within or out of a given circle, let a straight line AC be drawn to the circumference, in which take AB, so that AB. AC may always be equal to a given space; find the locus of the point B.

ANALYTICAL GEOMETRY

OF

THREE DIMENSIONS.

EQUATIONS OF A POINT.

We have seen that the position of a point in a plane is determined when we know its distances from two straight lines drawn in that plane; in like manner we shall now proceed to show that the position of a point in space, is determined by its distances from three planes.

P

A

X'

X

N

Let there be three planes YAZ, XAZ, XAY, which we shall suppose to be perpendicular to each other, and whose intersections are the three straight lines AZ, AY, AX, each of which is perpendicular to the other two according to the principles established in the Geometry of Planes. Let us call the distances of a point in space from these three planes a, b, c, and let us suppose these distances are known, then the position of the point will be completely determined, provided that we have ascertained in the first instance, that the point is situated within the trihedral angle AXYZ.

For, take on the three straight lines AX, AY, AZ, the distances AN, AO, AQ, respectively, equal to a, b, c; through the points N, O, Q, draw planes parallel to the given planes.

Since the two first parallel planes have all their points situated at the distances a and b, respectively, from the planes YAZ, XAZ, it follows that all the points of the straight line PM, which is the common intersection of these two planes have exclusively the property of being at the same distances from the planes YAZ, XAZ. Hence the point sought must be situated in the straight line PM. Again, the point sought must be situated somewhere in the third plane PnQo which is parallel to XAY, since all the points in this plane have exclusively the property of being at the distance c from the plane XAY. Hence the point sought must be the point P in which the third plane cuts the common intersection of the two first, and thus its position is altogether determined. We may designate by x the distance of a point from the plane YAZ reckoned along AX;

U U

We may designate by y the distance of a point from the plane XAZ reckoned along AY;

We may designate by z the distance of a point from the plane XAY reckoned along AZ.

So that AX, AY, AZ, the intersections of the three planes, two and two, will be the axes of x's, of y's, and of z's. They are called conjointly Axes of Coordinates, the three planes the Co-ordinate Planes, and the three distances the Co-ordinates of a point. These terms are all analogous to those already employed in Analytical Geometry of two dimensions.

The plane YAZ perpendicular to the axis of x's, is called the plane yz; The plane XAZ perpendicular to the axis of y's, is called the plane xz ; The plane XAY perpendicular to the axis of z's, is called the plane xy. This last plane is usually represented in a horizontal position, and the two others in a vertical position.

It follows from what has been said above that the equations

x = a, y = b, z = c

(a, b, c being known quantities) are sufficient to determine the position of a point in space, they are for that reason called the Equations of a point in space. We must remark, that, since the three co-ordinate planes v hen prolonged indefinitely determine eight trihedral angles, viz. four formed above the plane of xy, and four formed below the same plane; it is necessary for us to express analytically in which of these eight angles the point is situated. It is sufficient for this purpose, to extend to planes the principles which have been applied to distances from points and straight lines, that is to say, if we regard as POSITIVE distances reckoned along AX to the right of A, we must regard as NEGATIVE distances reckoned along AX to the left of A, that is to say, in the direction AX, the remark applies to the two other co-ordinate axes.

We must therefore consider in the quantities a, b, c, not only the numerical value of these quantities, but also the signs with which they are affected, in order that we may be enabled to determine in which of the eight trihedral angles about the point A the required point is situated.

According to this principle we have, in order to express completely the position of a point in space, the following combinations:

x = + a, y = X=- a, y = x = +a, y = x = + a, y = x=- a, y = x=- a, y = X = +a, y =

+ b, z = + c,
+ b, z = +c,
— · b, z=+c,
+ b, z = — c,

point situated in the angle AXYZ, point situated in the angle AXYZ, point situated in the angle AXY'Z, point situated in the angle AXYZ', +c, point situated in the angle AX'Y'Z, + b, z = — c, point situated in the angle AXYZ, b, z= c, point situated in the angle AXY'Z, a, y = b, z=- c, point situated in the angle AX'Y'Z',

.b, z=

in all, eight combinations, viz. two systems in which the signs are the same, three in which one sign is negative and the two others positive, and three in which one sign is positive and the two others negative.

There are also some particular positions of the point which it is proper to notice. For example, in order to express that a point is situated in the plane xy, we must write that its distance from that plane is nothing, and we shall have for the equation of such a point

x = a, y = b, z = 0.