We might, in like manner, have given a purely algebraical proof of Ex. 3.

Hence, also, if the points be joined to the centre where two parallel tangents meet any tangent, the joining lines will be conjugate diameters.

Ex. 5. Given, in magnitude and position, two conjugate semidiameters, Oa, Ob, of a central conic, to determine the axes. W

A

α

Р

M

B

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), 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 = Oa. 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 semidiameters, 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.

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

Forming, by Art. 40, the equation of a line drawn through

c2 being used, as in Art. 164, to denote a2 - b2.

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

P

M A

N

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 b = 0. The intercept CN, therefore, is constantly = 0 in the case of the circle, or every normal to a circle passes through its centre.

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

c2 a2

b2

a2

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.

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

But if be the semidiameter conjugate to CP, the quantity within the parentheses (Art. 177). Hence the length of the

=

If the normal be produced to meet the axis minor it can be proved, in like manner, that its length. Hence, the rectangle under the segments of the normal is equal to the square of the semiconjugate diameter.

Again, we found (Art. 181) that the perpendicular from the

ab

centre on the tangent= Hence, the rectangle under the normal

and the perpendicular from the centre on the tangent, is constant and equal to the square of the semiaxis.

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

Ex. 1. To draw a normal to an ellipse or hyperbola passing through a given point. Let the point on the curve at which the normal is drawn be XY, then its equation

and if this normal passes through a fixed point x'y', we have the relation

Hence 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

(for F

Ax2+Bxy + Cy2+ Ey = 0

0, because the origin is on the curve, and D = 0 (Art. 132), because the tangent is supposed to be the axis of x, whose equation is = 0). Now, let the equation of any two lines through the origin be

x2 + pxy + qy2 = 0.

=

Multiply this equation by A, and subtract it from that of the curve, and we get (B - Ap) xy + (C − Aq) y2 + Ey = 0.

This (Art. 36) is the

equation of a figure 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.

E

=

Aq - C

The point where this chord meets the normal (the axis of y) is y lines are at right angles q stant length

; but if the 1 (Art. 70), and the intercept on the normal has the con

==

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

E
Aq - C

is constant.

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

The co-ordinates of the intersection of the lines

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

where X, Y are the co-ordinates of the intersection of tangents, found in the last Example.

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

To express the distance of any point on an ellipse from the focus. Since the co-ordinates of one focus are (x = + c, y = 0), the square of the distance of any point from it

= (x′ − c)2 + y2 = x2 + y2 − 2cx' + c2.

=

But (Art. 177)

Hence

[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.]

Y

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 formulæ. FP+ FP= 2a,

Hence

or, The sum of the distances of any point on an ellipse from the foci is constant and equal to the axis major.

187. In applying the preceding proposition to the hyperbola, we obtain the same value for FP2; but in extracting the square root we must change the sign in the value of FP, for in the hyperbola a is greater than a, and e is greater than 1.

Hence, a – ex is constantly negative; the absolute magnitude, therefore, of the radius vector is

Therefore, in the hyperbola, the difference of the focal radii is constant, and equal to the transverse axis.

For both curves the rectangle under the focal radii a2 - e2x2, that is (Art. 177), is equal to the square of the semiconjugate dia

meter.

188. The reader may prove the converse of the above results by seeking the locus of the vertex of a triangle, if the base and either sum or difference of sides be given.

Taking the middle point of the base (= 2c) for origin, the equation is

√ {y2 + (c + x)2} ± √ {y2 + (c − x)2}· which, when cleared of radicals, becomes

=

2a,

Now, if the sum of the sides be given, since the sum must always be greater than the base, a is greater than c, therefore the coefficient of y' is positive, and the locus an ellipse.

If the difference be given, a is less than c, the coefficient of y is negative, and the locus an hyperbola.