y= 2r sin. 0 r sin. 20 From the former we easily find

Cos. = 1±√ 3r−2x which gives

stituted in the latter we get

3r + 2x + 2 √3r-2, which being sub

2

× 1 = √ 3r−2x, which, after proper reductions, will give

2

(x2 + y° — rx)2 = r2 { (x − r)® + y2}.

104. Let the part produced of the given line a; then by the question

105.

2

Let AB (Fig. 58,) = a, be the given line, required

to construct a Tetraedron upon it.

In the plane ABC construct the equilateral ▲ ABC, and let O be the point in it equally distant from A, B, C, and make OD 1

plane ABC = a

2; then AD, BD, CD, being joined, the

3

figure ABCD will be the Tetraedron required.

For AO BA :: sin. 30°: sin. 60°

.. the four faces are equal equilateral ▲ &c.

Again, r being taken in OD so that rArD, we have AO2 + Or2 = Ar2 = 1D2 = A02 + (OD − rD)2 = AO2 + OD2 + rD 20D × rD.

a2

.. 2 OD × rD = A02 + OD2 = AD2 = a2 or rD =

=

a2

20D 2a√3/

scribing sphere.

a

=

3

which is the radius of the circum

106.

To and the focus of a given parabola, either use the method adopted in Prob. 95, or the following one; drawing two parallel lines in the parabola QV, Q'V', and let them be bisected ¡n V, V', join VV' and produce it to meet the curve in P, thus determining an abscissa PV, corresponding to the ordinate QV. In like manner drawing any two parallel lines not parallel to the former, we get another pair of oblique co-ordinates pv, qv.

Now by property of the curve, we have (S being the focus)

with P, p, as centres, and radii SP, Sp, we describe two circular arcs, their intersection will be the focus required.

107.

From similar A OPM, ONQ, and the property of

the parabola, we have

AN AM :: QN2: PM2 :: ON2: OM2

;: (AO — AN) : (AM - AO)2

2AO

.. AN × (AM2 - 240 × AM + AO') = AM × (AN2 –

2AO × AN + AO2), and by reduction

AO2 × (AM -AN)

.. AO

AN

AN. AM (AM- AN)

AN × AM, or

108.

Let AP (Fig. 47,) be any conic section whose focus is S, and vertex A; also let La, Aa, be the tangents at the extremity of the latus rectum SL, and vertex A intersecting in a; then Aa

AS.

b

For supposing y=

√2ax = x2 to be the equation to the conic

α

section, (which will be an ellipse, parabola, or circle, when the negative sign is taken, and an hyperbola when the positive,) a and b being axes, and x, y, originating in A, we have.

(x' — x) ..... (1) the equation to the tangent

But a2 = b2 + (a ‡ m)2 = b2 + a2 + m2 + 2am from a wellknown property of the curve,

109.

If R, r, be the radii of the circles, circumscribing

and inscribing a A whose sides are a, b, c, then

For, supposing A, B, C, the angles to which a, b, c, are respectively opposite, since those angles are each bisected by the lines joining them and the centre of the inscribed O, we have

Again, since the chord of an arc = 2 sin. of half that arc, we

(a+b+c).(a+b−c)(c+b−a). (c+a−b)

110.

The equation to an equilateral hyperbola whose

axes are unity, and co-ordinates at the centre being

y = √1+x2

1. (x + √1+x2) = n. l. (x' + √1+x2)

:. x + √ 1+x2 = (x2 + √ 1+x'3)"

111.

1

1+

: √1+x2

√1+x2

=

=

=

x + √ 1 + x2

1

=

(1);

Let LP (Fig. 47,) be a tangent at L the extremity of the latus rectum of the ellipse, meeting any ordinate PN produced in P'; then S being the focus, required to shew that SP = NP'.

a and b being the semi-axes of the ellipse, y, x, its co-ordinates at the vertex, and y', x', those of the tangent at any point, originating also at the vertex, we have

x'2 2 √ a2b2 (a ± √ a2 − b3) x2 +

2a2b2 + 2a√ a2 — b3