either case would be plus. It is evident from the equation y2 =

b2

a2(x2

a2) that x may be taken any quantity not numerically less than a, plus or minus; and that for every value of x there will be two values of y, numerically equal, but of contrary signs. Also, for every value of y there will be two values of x, numerically equal, but of contrary signs. Hence each axis bisects every line drawn parallel to the other axis, having its extremities in the curve of the hyperbola, or the opposite hyperbolas.

b2

In the expression y=√(x2

a

a2), x cannot be less than a, other

wise we would have the square root of a minus quantity, which is impossible.

-225

Draw the axes XX', YY', intersecting in O, and take OA, OB, each 5, on opposite sides of O; then A and B will be the points in which the curve represented by the equation 25y9y2: cuts the axis of x. Lay also OC=+3, and OD=-3. Now, as x cannot be less than 5, take x ±6, x ±7,x= ±8, x=9, etc., and lay these values of x both ways, from O, on the axis XX', to 6, 7, 8, 9, etc., and through these points, respectively, draw lines parallel to YY'. If x=6, y = ± ±√11=±1.99, which lay on the parallels through 6, up and down, to a, a and a', a', and these will be points in the curve.

3

3

If x 7, y

±√24±2.94, which lay on the parallels

5

through 7, up and down, to b, b and b', b', and these will be other points in the curve.

3

If x 8, y ±√39

±3.71; if x=

9, y

=±4.49; if x=

10, y=5.20, etc., which lay on the corresponding parallels through 8, 9, etc. to c, c, d, d, etc., and to c', c', d', d', etc., and then each of the curves passing through the points A, a, b, c, d, etc. both ways from A, and through B, a', b', c', d', etc. both ways from B, will be a hyperbola, and the two together are called opposite hyperbolas.

O is the centre, AB the first axis, CD the second axis, and A and B are called the vertices of the hyperbolas.

As equal ordinates, or values of y, correspond to equal abscissas, or values of x, the opposite hyperbolas are similar figures.

Join BC, and lay the distance BC, from 0, on the axis of x, both ways, to F and F"; then these points are the foci of the hyperbola. Parallel to CD draw any double ordinate HGE, H'G'E'; then, of the point H, GH is the ordinate, and BG and AG the two corresponding abscissas; and of the point H', G'H' is the ordinate, and AG', BG' the two corresponding abscissas.

Scholium.-Join AD, and it will be parallel to BC, because the alternate angles OBC and OAD are equal.

Example 2.-Determine the axis, and construct the hyperbola, whose equation is y2-3x2-5.

25

3

and the equation is of the form a2y'-b2x2-a2b2, which

is the equation of a hyperbola. We have for the semi-axes, a

If x 12, y =±2.05; if x=

=2, y

1.29, which lay on the axis of x from O to A and B, and these will be the vertices of the opposite hyperbolas. On the axis of y lay also OC and OD each =√5 52.24. ±2.65; if x=212, y 15

3.71. It is evident that x cannot be numerically less than

√ 3

Lay 01, 02, 03, etc., the values of x, on XX', to the right of O, and also to the left of O, to 1, 2, 3, etc., and through the points 1, 2, 3, parallel to CD, draw lines, on which parallels lay the corresponding values of y to a, b, c, etc., and to a', b', c', etc., and these give points in the required curve, any number of which may be obtained by assuming different values for x. Each of the curves

-5 1X-3

* It will be seen that the multiplier is the absolute quantity -5 divided by the product of the co-efficients of y2 and x2, with their respective signs.

passing through the points A, a, b, c, etc., both ways from A, and through B, a', b', c', etc., both ways from B, will be a hyperbola, of which the equation is y2 - 3x2

opposite hyperbolas.

2

-5. The two together are called

Join AD and BC, and they will be parallel. Lay BC on the axis of x, from O, both ways, to F and F", which points will be the foci, AB is the first axis, and CD the second. If the double ordinates HGE and H'G'E' be drawn, HG is the ordinate and AG and BG the abscissas of the point H; and H'G' the ordinate and AG' and BG' the abscissas of the point H'.

Example 3.-Determine the axes, and construct the hyperbola, whose equation is 2y2-4x24. Multiply the given equation by -2, where a2 1, b2

4

2 x -4

1,* and we get —y2 + 2x2

2, a2b2 ——2, and the given equation is reduced to the form a2y2b2x2-a2b2, which is the equation of a hyperbola. The first axis 2√1, which shows that the curve cannot meet the axis of x. If y 0, x=1; if x =0, y =±√2

1.41. Lay OA, OB each equal to 1; then AB the first axis. Lay OC and CD each equal to /2= 1.41; then CD the second axis. Now, y2 2x2 + 2; hence y 2x2+2. The least value of y 0; then y

is when x

G

E

H

14.

d

'd

13

12

B

N

1.41=OC or OD; x2

A

13

If y = 11, x ==

2y2 — 4.

±.35; if y 2, x = ±1.00; if y = 21, x = =±1.46; if y = 3, x = ±1.87.

Lay the values of y from O, up and down, to 1, 2, 3, 4, etc., and through these points draw lines parallel to the axis XX', and on these parallels, from the points 1, 2, 3, 4, etc., lay the corresponding values of x to a, a; b, b; c, c; d, d, etc., and to a', a'; b', b';`` c', c'; d', d', etc.; then each of the curves passing through the points C, a, b, c,

See foot-note to last exan ple.

d, etc., both ways from C, and through D, a', b', c', d', etc., both ways from D, will be a hyperbola, whose equation is 2y2-4x2 = 4.

Join AC and DB, and they will be parallel. Lay AC on the axis of y, both ways from O, to F and F", which points will be the foci. C and D are the vertices of the opposite hyperbolas.

If parallel to XX' the double ordinates HGE and H'G'E' be drawn, HG will be the ordinate and CG and DG the abscissas of the point H; and H'G' the ordinate and C'G' and DG' the abscissas of the point H'.

Construction.-Draw PAV, and EF at right angles to it at A. In PAV take any point P below EAF, and V above EAF. Draw any number of lines, as Pb, Pb, etc., on each side of PV, and make extensions beyond EF from P, as ab, ab, etc., each equal to AV, and the curve passing through the point V, and the several points b on each side of V, will be the superior conchoid.

Now, on the same lines drawn from P, lay the same distance equal to AV below EAF, as AV', ab', ab', etc., on each side of AP, and the curve passing through the point V', and the several points b' on each side of V', will be the inferior conchoid.

The point P is called the pole; the line EAF, the directrix; V, the vertex of the superior conchoid; and V', of the inferior.

Property 1.-The directrix EAF is an asymptote to both curves. For, let fall the perpendiculars bG, b'G' on the directrix, and bH, b'H' on PAV. Then the triangles baG, PaA, taking Pab the line from the remote extremity of which the perpendiculars bG and bH are let fall, are similar, and we have Pa: PA :: ab (AV): bG. Hence

* For a treatise on the interesting subject of curves, see Prof. Leslie's Geometry of Curve Lines, or, the "Application of Algebra to the Doctrine of Curves," at the close of the second volume of Bonnycastle's Algebra,