P

M

B

a

The following construction is founded on the theorem proved in the last Example:-Through a the extremity of either diameter, draw a parallel to the other; it must of course be a tangent to the curve. Now, on Oa take a point P, such that the rectangle Oa.aP Ob2 (on the side remote from O for the ellipse, on the same side for the hyperbola), A and describe a circle through O, P, having its centre on aC, then the lines OA, OB are the axes of the curve; for, since the rectangle Aa.aB = 0a.aP = Ob2, the lines OA, OB are conjugate diameters, and since AB is a diameter of the circle, the angle AOB is right.

Ex. 6. Given any two semi-diameters, if from the extremity of each an ordinate be drawn to the other, the triangles so formed will be equal in area.

Ex. 7. Or if tangents be drawn at the extremity of each, the triangles so formed will be equal in area.

THE NORMAL.

180. A line drawn through any point of a curve perpendicular to the tangent at that point is called the Normal.

Forming, by Art. 32, the equation of a line drawn through

(x'y') perpendicular to

of the normal to a conic

or

xxyy
b3

+ 1),

we find for the equation

c' being used, as in Art. 161, to denote a3 — b3.

Hence we can find the portion CN intercepted by the normal

on either axis; for, making y = 0 in

the equation just given, we find

B

We can thus draw a normal to

an ellipse from any point on the axis,

for given CN we can find x', the abscissa of the point through which the normal is drawn.

་

The circle may be considered as an ellipse whose eccentricity =0, since c2=a-6-0. The intercept CN, therefore, is constantly = 0 in the case of the circle, or every normal to a circle passes through its centre.

181. The portion MN intercepted on the axis between the normal and ordinate is called the Subnormal. Its length is, by the last Article,

x =

The normal, therefore, cuts the abscissa into parts which are in a constant ratio.

If a tangent drawn at the point P cut the axis in T, the intercept MT is, in like manner, called the Subtangent.

Since the whole length CT= (Art. 169), the subtangent

х

The length of the normal can also be easily found. For

But if b' be the semi-diameter conjugate to CP, the quantity within the parentheses 6" (Art. 173). Hence the length of the

bb'

a

H

If the normal be produced to meet the axis minor, it can be

angle under the segments of the normal is equal to the square of the conjugate semi-diameter.

Again, we found (Art. 175) that the perpendicular from the

normal and the perpendicular from the centre on the tangent is constant and equal to the square of the semi-axis minor.

Thus, too, we can express the normal in terms of the angle it makes with the axis, for

Ex. 1. To draw a normal to an ellipse or hyperbola passing through a given point. The equation of the normal, a'x'y — b2x'y = c2x'y', expresses a relation between the coordinates 'y' of any point on the curve, and ry the coordinates of any point on the normal at x'y'. We express that the point on the normal is known, and the point on the curve sought, by removing the accents from the coordinates of the latter

point, and accentuating those of the former. Thus we find that the points on the curve, whose normals will pass through (x'y') are the points of intersection of the given curve with the hyperbola

Ex. 2. If through a given point on a conic any two lines at right angles to each other be drawn to meet the curve, the line joining their extremities will pass through a fixed point on the normal.

Let us take for axes the tangent and normal at the given point, then the equation of the curve must be of the form

ax2 + 2hxy + by2 + 2fy = 0,

(for c = 0, because the origin is on the curve, and g = 0 (Art. 144), because the tangent is supposed to be the axis of x, whose equation is y = 0).

Now, let the equation of any two lines through the origin be

x2+2pxy + qy2 = 0.

Multiply this equation by a, and subtract it from that of the curve, and we get

2 (hap) xy + (b − aq) y2 + 2fy = 0.

This (Art. 40) is the equation of a locus passing through the points of intersection of the lines and conic; but it may evidently be resolved into y = 0 (the equation of the tangent at the given point), and

which must be the equation of the chord joining the extremities of the given lines.

aq

2f

b; but if

The point where this chord meets the normal (the axis of y) is y = the lines are at right angles q = - 1 (Art. 74), and the intercept on the normal has the constant length

2f =

a+b

*

If the curve be an equilateral hyperbola, a + b = 0, and the line in question is constantly parallel to the normal. Thus then, if through any point on an equilateral hyperbola be drawn two chords at right angles, the perpendicular let fall on the line joining their extremities is the tangent to the curve.

Ex. 3. To find the coordinates of the intersection of the tangents at the points x'y', x"y".

The coordinates of the intersection of the lines

Ex. 4. To find the coordinates of the intersection of the normals at the points x'y', x"y".

Ans.

(a2 — b2) x'x'X

x=

(b2 - a2) y'y" Y

, y=

* This theorem will be equally true if the lines be drawn so as to make with the normal angles the product of whose tangents is constant, for, in this case, q is constant, and, therefore, the intercept is constant.

2f aq-b

where X, Y are the coordinates of the intersection of tangents, found in the last Example.

The values of X and Y may be written in other forms. Since by combining the equations

181 (a). Let CP, CQ be a pair of conjugate semi-diameters

Hence CD' = (a+b). Similarly for CD. The axis-major bisects the angle DCD.

D'N=D'P+PN=b' +

Similarly DN=

=-
a

bb'

a

=

For the line

(a + b).

a

(a - b). At the point N, therefore, the

base of the triangle DCD' is divided in the ratio of the sides, and, therefore, CN is the internal bisector of the vertical angle. In like manner, it is proved that CN' is the external bisector.

Hence then, being given two conjugate semi-diameters CP, CQ in magnitude and position, we are given the axes in magnitude and position. For we have only from P to let fall on CQ the perpendicular PR; to take PD, PD each equal CQ; then the axes are in direction the bisectors of the angle DCD; while their lengths are the sum and difference of CD, CD'.

THE FOCI.

182. If on the axis major of an ellipse we take two points

equidistant from the centre whose common distance

T

R

T

The foci of a hyperbola are two points on the transverse axis, at a distance from the centre still = c, c being in the hyperbola

= √√/(a*+b").

To express the distance of any point on an ellipse from the focus.

Since the coordinates of one focus are (x=+c, y=0), the square of the distance of any point from it

[We reject the value (ex′ – a) obtained by giving the other sign to the square root. For, since x is less than a, and e less than 1, the quantity ex-a is constantly negative, and therefore does not concern us, as we are now considering, not the direction, but the absolute magnitude of the radius vector FP.] We have, similarly, the distance from the other focus

F'P=a+ex',

-

since we have only to write - c for + c in the preceding formula. Hence FP+F'P=2a, or, The sum of the distances of any point on an ellipse from the foci is constant, and equal to the axis major.

183. In applying the preceding proposition to the hyperbola, we obtain the same value for FP'; but in extracting the square

A A