180 EVERY CURVE OF THE SECOND DEGREE IS A CONIC.

The equation (i) will become

a (x cos y sin 0)2 + 2h (x cos 0 — y sin 0) (x sin 0 + y cos 0) +b(xsine+ycos)2+2g (x cos 0—ysin 0)+2ƒ (x sin0+ycose) .(ii).

+c=0

The coefficient of xy in (ii) is

2 (b− a) sin cos 0 + 2h (cos30 — sin30);

and this will be zero, if

any

tan 20 =

2h a-b

...(iii).

Since an angle can be found whose tangent is equal to

real quantity whatever, the angle ( = 1 tan ̃1

all cases real.

Equation (ii) may now be written

2h

a-b

Ax+By+2Gx+2Fy+ C=0................(iv).

is in

If neither A nor B be zero, we can write equation (iv)

in the form

If the right side of (v) be zero, the equation will represent two straight lines [Art. 35].

If however the right side of (v) be not zero, we have

the equation

which we know represents an ellipse if both denominators are positive, and an hyperbola if one denominator is positive and the other negative.

If both denominators are negative, it is clear that no real values of x and of y will satisfy the equation. In this case the curve is an imaginary ellipse.

Next let A or B be zero, A suppose. [A and B cannot both be zero by Art. 53.] Equation (iv) can then be written

If G= 0, this equation represents a pair of parallel straight lines.

If G be not zero, we may write the equation

which represents a parabola, whose axis is parallel to the axis of x.

Hence in all cases the curve represented by the general equation of the second degree is a conic.

168. To find the co-ordinates of the centre of a conic. We have seen [Art. 109] that when the origin of coordinates is the centre of a conic its equation does not contain any terms involving the first power of the variables. To find the centre of the conic, we must therefore change the origin to some point (x, y), and choose x, y', so that the coefficients of x and y in the transformed equation may be zero.

Let the equation of the conic be

ax2+2hxy+by2 + 2gx + 2fy + c = 0.

The equation referred to parallel axes through the point (x, y) will be found by substituting x + x' for X, and y+y for y, and will therefore be

a (x+x)2 + 2h (x + x') (y + y) +b (y + y)2 + 2g (x+x')

+2f (y + y) + c = 0, or ax2+2hxy + by2 + 2x (ax' + hy' + g) + 2y (hx' + by' +ƒ). +ax2+2hx'y' + by'2 + 2gx' + 2ƒy' + c = 0. The coefficients of x and y will both be zero in the above, if x and y' be so chosen that

and

ax' + hy' + g=0...

hx + by +ƒ=0.......

.(i),

.(ii).

The equation referred to (x, y) as origin will then be

ax2+2hxy + by2 + c' =0.....

..(iii),

where c' = ax2 + 2hx'y' + by'2 + 2gx' + 2ƒy' + c......(iv). ` Hence the co-ordinates of the centre of the conic are the values of x' and y' given by the equations (i) and (ii). The centre is therefore the point

When ab-h2 = 0, the co-ordinates of the centre are infinite, and the curve is therefore a parabola [Art. 157]. If however hf - bg= 0 and ab - h2 = 0; that is, if

the equations (i) and (ii) represent the same straight line, and any point of that line is a centre. The locus in this case is a pair of parallel straight lines.

In the above investigation the axes may be either rectangular or oblique.

Subsequent investigations which hold good for oblique axes will be distinguished by the sign (w).

169. Multiply equations (i) and (ii) of the preceding Article by x, y respectively, and subtract the sum from the right-hand member of (iv); then we have

• 170. The expression abc +2fgh-af-bg-ch is usually denoted by the symbol A, and is called the discriminant of

ax2+2hxy + by2+2gx+2fy+c.

A=0 is the condition that the conic may be two straight lines.

For, if A is zero, c' is zero; and in that case equation (iii) Art. 168 will represent two straight lines.

This is the condition we found in Art. 37.

(w).

171. To find the position and magnitude of the axes of the conic whose equation is ax + 2hxy+by2 = 1.

If a conic be cut by any concentric circle, the diameters through the points of intersection will be equally inclined to the axes of the conic, and will be coincident if the radius of the circle be equal to either of the semi-axes of the conic.

Now the lines through the origin and through the points of intersection of the conic and the circle whose equation is x2 + y2 = r2, are given by the equation

(a) a2 + 2hxy + (b ——) y2 = 0.............(i).

These lines will be coincident, if

and they will then coincide with one or other of the axes of the conic.

Hence the lengths of the semi-axes of the conic are the roots of the equation (ii), that is of the equation

Hence if we substitute in (iv) either root of the equation (iii) we get the equation of the corresponding axis.

In the above we have supposed the axes to be rectangular. If however they are inclined at an angle w the investigation must be slightly modified, for the equation of the circle of radius r will be x2 + 2xy cos w + y2 = r2.

172. To find the axis and latus rectum of a parabola.

If the equation

ax2+2hxy + by2+2gx+2fy+c=0

represent a parabola, the terms of the second degree form a perfect square. [This follows from the fact that the equation of any parabola can be expressed in the form y-4a'x0, and therefore with any axes the equation will be of the form

(lx+my+ n)2 - 4a' (l'x + m'y +n') = 0.] Hence the equation is equivalent to

.........

(ax+ By)2 + 2gx+2fy + c = 0 .........................(i), where a2= a, and ß2= b.

From (i) we see that the square of the perpendicular on the line ax + By = 0 varies as the perpendicular on the line 2gx+2fy+c=0. These lines are not at right angles, but we may write the equation (i) in the form

(ax+By+λ)2 = 2x (λa − g) + 2y (λß − ƒ) +λ3 — c,

and the two straight lines, whose equations are

ax+By+λ = 0, and 2x (λa − g) + 2y (λß −ƒ) + λ2— c = 0, will be at right angles to one another, if

ax+ By +λ=0 and 2(aλ − g) x + 2 (Bλ −ƒ) y +λ2 — c = 0 for new axes of x and y respectively, and we get

y2 = 4px,

and this we know is the equation of a parabola referred to its axis and the tangent at the vertex.

To find the latus-rectum, we write the equation in the form