36.

Let AB, Dd (Fig. 27,) be any two diameters of the O ADBd one another, and let Dd be indefinitely produced both ways to E and e. Then, any chord AF being drawn intersecting Dd in P, and PM = cos. FB being erected Dd, required the locus of the point M.

Let PC = x, PM y, and AC = r. Then, from similar A we get

Let x = 0. Then yr, or the curve passes through B.

Let xr. Then y = 0, or the curve passes through

Let x 0. Then tan. = 0, or = 0, or the tangent at B

is parallel to CE.

Let xr. Then tan. = 1, or = 135°, or = 45°, ..the tangent at D is inclined to CE at an angle of 135°, and at d at an angle of 45°.

Hence, if CB be bisected in G, and Ff be drawn through G AB meeting the locus in M, m ; M, m will be points of contrary flexure, and the locus will be as it is described in the figure.

37.

To find the relation between the ordinate and abscissa of a curve (y, x) from the equation,

e√√dx2+dy2 = e2

Differentiating, we have

S√dx2

√ dx2 + dy2. e √ √ dx2+dy2 = dx . ( e2 + e~')

38.

PN (Fig. 28,) is any ordinate of the ellipse ABɑ, and Q is a point in it such, that the line CQ joining it and the

centre of the ellipse C, shall always = PN; required the locus of Q.

Let CN= x, QN = y, AC = a, and BC = b.

Then y2 + x2 = QC2 = PN2 =

a2

b2

(a2x2) by the equation

ab

√ a2 + b2

1

1

tion to an ellipse, whose semi-axes BC, CN'are b and

respectively.

The point N' always falls within the given ellipse, and may be found geometrically, by bisecting the ACB by the line CP', and drawing the ordinate P'N'.

39.

The problem may be generalized thus; ACB (Fig. 29,) being any given curve, whose equation is y'= f.x', and CD a straight line of given length, one of whose extremities is always in the line of abscissæ, and the other in the curve, then CD having every possible situation, required the locus of the point P which divides CD in a given ratio.

Let O be the origin of abscissæ of both the curve and the locus, ON = x', CN = y', OM = x, PM=y, CD = a, and PD CD

=

α

presses the relation between x and y in the locus generally.

y

r2

Hence Sydz = y √y + √ - - sin.

2y

+C; and transferring the origin of abscissæ

r2

-y2 = area 6PM, the correction in this case being = 0.

If ACB be a straight line, it will be found, in like manner that the locus is a conic section.

If our limits would permit, we might here resolve the inverse problem; given the equation to the locus of P, to find that of the extremity C; or, more generally, given the equations to the loci of two points of a straight line, to find that of any other point.

40. To trace a. (y — b)2 = x. (x — a)2, let y — b = u, and we get

Let x = 0.

u = ± (x − a) √x

a

Then u±0, and the curve passes through the point A (Fig. 30), the origin of u.