XIV. QUESTION 384, by PALABA.

Find the equation of the curve of which this is the property; if from a fixed point in the axis a perpendicular be drawn to it, and produced to meet a tangent to any point in the curve, the length of this perpendicular and tangent together, shall be double the length of the curve between the vertex and the point from which the tangent was drawn.

FIRST SOLUTION, by PALABA, the Proposer.

Constructing the figure, let

BA a, PC, CPY, AP

4

B

yx

x ::

xyy

*√(x2 + j2) = √(x2+ÿ1). Assumeỷ=px.. ÿ = p x

xpp

√(1 's p2)= *√(1+p2).•.

y

.. L. x + L. (p + √(1 + p2)) — L. √(1+p2) = c,

tan 4 OPC o, when x = a, ... L. a+L

x

p

(20

(4a — x) · √(x - a), which wants no cor

SECOND SOLUTION, by Mr. W. WALLACE, R. M. College.

Let BPC be the curve, в a point in which it meets the axis, A the fixed or given.

point in the axis, AF the perpendicular, and PD a tangent at any point P, meeting the perpendicular in D. Let Po be any ordinate, and DH a

perpendicular from

D to PQ.

Put AB = a, BQ = x, PQ = y, arch BP = z.

By the theory of curves,

also dr: dz :: DH: DP.

therefore q✔(1+p'); the letters p and q being now sub

stituted in the foregoing equation, it becomes

y + (a + x) (q —p) = 22.

Hence, taking the fluxions, we get,

..(A)

dy + (a + x) (dq — dp) + (q − p) dx = 2dz.

In this equation, substitute pdr for dy and qdx for dz, then, after reduction, it becomes,

(a + x) (dq-dp) = qdx;

dq

9

dp
√(1 + p2)

This equation by integration gives

log (a + x) = log √(1 + p2) — log { p + √(1+p3) } + logc;

and hence a + x =

c√(1+p2)

p + √(1 + p2)

.(B).

To determine the value of the arbitrary quantity c, we must recur to equation (A), and observing that x, y and z begin together, it there appears that when x, and consequently y and z are each o, then a(q- p) = o, or q = √(1 + p2) = ?, hence, 1+ p2 = p2, which shews, that when xo, then pis ср = 2p

infinite and in this case, equation (B) becomes a =

hence c = 2a, and the adjusted equation is

and taking the fluents, so that x and y may begin together, we

find

Sadx

1

y=

eva (2a√/x — ;x3);

and, after proper reduction,

gay2x (3a- x)'.

This is the equation of the curve, which appears to be a parabolic line of the third order, the 68th species according to Newton.

To investigate a geometrical construction for any point in the curve, let us take BC= 3BA 3a, then cQ = za · equation of the curve gives

3BC X PQ2

BQ × QC2.

x, and the

Draw CP, producing it to meet BL a perpendicular to BC in

K, and draw KI perpendicular to CK, meeting AB in I; then, by similar triangles and the property of a right-angled triangle,

cq2: QP2 :: CK2: KI2 :: CB: BI.

But, by the equation of the curve,

CQ: QP 3BC: BQ

BC: +BQ,

therefore BQ3B1, also cQ = 3A1. This property indicates an easy method of finding the point in which any ordinate PQ meets the curve, namely, by taking BIBQ, and describing a circle on IC as a diameter, to meet BL in K; a straight line drawn from c to K will manifestly pass through P the top of the ordinate.

This formula indicates an elegant property of the curve.

XV. QUESTION 385, by PALABA.

If the sine of incidence: sine of refraction :: 1:n, r and the radii of the surfaces, and t, its thickness, the distance (f) of the principal focus from the focal centre may be accurately determined from this expression,

SOLUTION, by PALABA, the Proposer.

Let AB be the lens whose axis is TR and centre E, R and r the centres of its surfaces, MS, Qa, PN a pencil of parallel rays incident upon it, of which Qabq passes through the centre; m and n the focal centres, also let PN be that ray which is incident perpendicularly on surface A. Take NV RV :: 1: n, and v is the focus after the first refraction. Join vr meeting the surface A in s and the line bq in G, then G is the focus of emergent rays (see Wood's Optics, 3rd Ed.). Then, from similar triangles, Rr: RV :: nr : no the distance of the principal focus from

Then.. NV: RV :: 1 : n, r: RV :: 1-7

(Wood's Optics) ... Er = TB —EB = r' —

-n)t

COR. If AB be a sphere, E is its centre, and m and n co