PROBLEM III.-To construct the parabola from its equation y2 = ax.* P Y d e α 1 12 13 14 5 6 C e B Analysis.-Let O be the origin of the axes XX', YY'. It is evident from the equation y2 =ax that if y = 0, x = 0; and if x 0, y = 0. Hence the line represented by the equation passes through the origin 0. It is evident, moreover, that a cannot be negative, or minus, for, if it were, ax would be minus, and we would have y -ax, which is impossible. Hence the curve must lie wholly on the right of the axis YY'. Also, since y Vax, every value of x will give two values of y, numerically equal, but of different signs: hence the axis of x bisects all lines drawn parallel to the axis of y and terminated at both ends by the curve; and it bisects, also, the area of any portion of the parabola cut off by a line parallel to the axis of y. Example 1.—In the equation y2 ax, take x 5, and it becomes y2 = 5x, or y 5x. By the table of square roots, annexed to √5±2.24; ±2.24; if x 2, y = √10 ±3.87; if x 4, y ±4.47; if x this volume, if x= 1, y ±3.16; if x 3, y 15 01=1, 5, y √25±5; if ŵ= 6, y 5.48, etc. Now, lay 01 02=2, 03=3, 044, etc., and through the points 1, 2, 3, 4, etc. draw lines parallel to the axis YY'; and, since all the values of y have both plus and minus signs, they must be laid both ways from the axis of x; lay la 2.24, 26—3.16, 3c 3.87, 4d 4.47, etc., laying each up and down, and the curve passing through O, a, b, c, d, etc., both ways from O, will be the parabola required. Example 2.-In the equation of the parabola, take x=-3, and it becomes y2 -3x, or y -3x. Here the values of x must be minus in order to have 3x plus and y2 plus. Hence the curve lies wholly to the left of the axis YY'. * See remark to Example 2, Analytical Geometry. proceed exactly as in Example 1, laying the values of x, 1, 2, 3, etc. to the left of YY', because all its values are necessarily negative, and hence the curve lies wholly to the left of YY'. NOTE.-By taking x=.5, 1.5, 2.5, 3.5, etc. in these examples, or other intervening quantities, intermediate points at pleasure may be obtained. THE ELLIPSE. PROBLEM I.-To construct the ellipse, having the axes given, and determine its equation. Construction.-Draw the axes AB and CD, bisecting each other at right angles at the point O, which will be the centre of the ellipse. Draw BF parallel and equal to OC or OD, and join OF. With the centre O and radius OA or OB, half the greater axis AB, describe the semicircle AEB, E being the prolongation of CD. Divide the semicircle AEB into six A C d C e e d a B 54 3 12 1 2345 S R 3 4 5 E 5 equal parts by laying the radius from A to 4, and from B to 4, and from E, both ways, to 2 and 2; then bisect each of these equal parts, and each quadrant will be divided into six equal parts in the points 1, 2, 3, etc., and the whole semicircle into twelve. Through each of these points draw lines parallel to CE, cutting the axis AB in the points 1, 2, 3, etc., and the line OF in the points S, R, c, Q, and P, the point Q coinciding with 2 on the semicircle. Now, take the distance 18 in the dividers, and lay it from each 5 which is on the axis, up and down, to a and a, giving four points in the ellipse. 17 and c. a b C e е d α b a B Take the distance 2R, in like manner, and lay it from each 4 on the axis, up and down, to b and b, giving four more points in the curve. In like manner, take the distance 3c (noting that c on OF is a point in the curve), and lay it from each 3 on the axis, up and down, to c Lay the distance 4Q from each 2 on the axis, up and down, to d and d; and lay 5P from each 1 on the axis, up and down, to e and e; then the curve passing through these twenty-four points A, B, C, D, and each of the points a, b, c, d, e, taken in regular succession from A through C, B, and D, will be an ellipse. It will be noticed that 1S is just as far, measured on the semicircle, from OE as 5a is from B or A; that is, E5 B1=A1; also, 2R as far from OE as 4b is from B or A, so that the arcs E5 — B1 or A1, E4 B2 or A2, E2 — B4 or A4, etc. Also, since B4 and E2 are equal, each being composed of four equal parts of the semicircle, we have the line 04 equal to the line 2d4. 2 4 3 2 1 1 2 3 45 Demonstration.-Taking any point in the curve, as d, in the quadrant BD, 2d, by construction, 4Q. In similar triangles OBF and 04Q, we have 04 (or 2d4): 4Q (or 2 Rd) :: OB: BF Hence (2d4): (2 Rd)2:: OB2: BF2 or OD. But (IV. 23, cor.*) (2d4)2== A2 × 2B; hence A2 × 2B: (2 Rd)2 : : OB2 : OD2; that is, as the rectangle of the two abscissas of the point d (A2 × 2B) is to the square of the ordinate (2Rd), so is the square of the semi-axis major (OB) to the square of the semi-axis minor (OD), which is the property of the ellipse. If OB=α, OD—b, 02 == x, and 2Rd У the ordinate, then A2 a+x, and 2B—a— x, the two abscissas, and the above proportion becomes (a + x) x (a-x): y2:: a2: b2, or a2x2: y2:: a2: b2; whence y2 = 2(a2 — x2), or y = -√ a2 — x2; or a2y2 + b2x2 -- a2b2, either of which is the equation of the ellipse. PROBLEM II.-Having an ellipse given, it is required to find its axes, foci, focal tangents, and to draw a tangent to any point in the ellipse. Construction. In the given ellipse, draw any two parallel chords, as EG and HI; bisect these in J and K, and through these points * VI. 13. draw the line LJKM, which will be a diameter to the ellipse. Bisect LM in O, which will be the centre of the ellipse. With the centre Through O draw COD perpendicular to AB, and it will be the minor axis. With Cas a centre and radius OA or OB describe arcs, cutting the axis AB in the points F and F", which will be the two foci of the ellipse. Through the centre O, parallel to either of the chords EG and HI, draw WOX, and it will be the conjugate diameter to LOM, and LM and WX are called a pair of conjugate diameters, of which each one bisects all chords in the ellipse drawn parallel to the other, such chords being called double ordinates to the diameter that bisects them, and the two parts into which the diameter is divided are the corresponding abscissas. The lines through the extremities of either diameter, parallel to its conjugate or double ordinate, are tangents to the ellipse at those points. Thus, MQ, drawn through M, parallel to the diameter WOX or the chord HI, is a tangent to the ellipse at the point M. On AB, with O as the centre, describe a semicircle. From the focus F, perpendicular to AB, draw the line FTV, draw VY a tangent to the semicircle, and join T'Y; then T'Y will be a tangent to the ellipse at T', and it is called the focal tangent. To draw a tangent to the ellipse at any given point, as Z, parallel to COD draw the double ordinate URZ, and produce it to meet the semicircle at S. At S draw a tangent ST to the semicircle, meeting the axis AB, produced, in T, and join 12, which will be the tangent to the ellipse at Z. The tangents to the semicircle and ellipse, from points in the same ordinate, meet at the same point in the axis. 2 Since, as shown in last problem, AR × RB: RZ2:: A02: OD2, and AR × RB = RS2, we have RS2: RZ2 :: AO2 : OD2, or RS: RZ:: A0: OD; that is, the ordinate of the circle at any point of the axis is to the ordinate of the ellipse at the same point, as AO to OD, or as the major axis to the minor axis, in a constant ratio. Hence these ordinates to the circle and ellipse will always be proportionate to each other, and we have SR: ZR::VF: TF, etc. PROBLEM III.—To find the area of an ellipse by the calculus. (See problem in the article "Calculus.") Put 40 =a, OC=b, OR—x, RU—y. Now, to find the area of the quadrant BOCUB of the ellipse, RU is the moving line, supposed to commence at OC, and moving parallel to itself through the distance OB, to generate the surface of the quadrant BOCB. b2 Now, by Problem I., the equation of the ellipse is y2 ~2(a2 —— x2), 2 dx 4 of the problem in the article "Calculus," ✔a2x2× dæ is the b area of a quadrant of a circle whose radius is a. Hence ✔ x dx b a × the area of the quadrant of a circle whose radius is a、 That is, we have a:b:: area of quadrant of circle: area quadrant of ellipse: area of circle: area of ellipse, the diameter of the circle being the major axis of the ellipse, 2a. Then the area of the circle 4a2 × .78539, and a:b::4a2 × .78539: area of ellipse 4ab x .78539 2a × 2b × .78539 AB × CD × .78539. That is, the area of an ellipse is equal to the product of the axes of the ellipse multiplied by the area of a circle whose diameter is unity, or it is equal to the area of a circle whose diameter is a mean proportional between the two axes, ✔4ab. PROBLEM IV.-To construct the ellipse from its equation y2 2 Analysis. It is evident from the equation ay2 + b2x2 = a2b3 that if y 0, x +a ora, which shows that the curve represented by the equation cuts the axis of x in two points, distant a from the origin, one to the right, the other to the left of 0. Also, if X 0, y +bor —b, showing that the line cuts the axis Y in two points, distant b from the origin, one above and the other below. of See remark to Example 11, Analytical Geometry. |