COR. 1.-A point is without, in, or within, the curve, according as its distance from the focus is greater, equal, or less, than its distance from the directrix.

COR. 2.-A perpendicular to the axis, at its vertex, is a tangent to the parabola.

In a

For HIVCVF, and VFFH, therefore HI FH; and hence (Cor. 1) H is without the curve. similar manner, every other point of VH may be shown to be without the curve; therefore VH is a tangent.

PROPOSITION II.

Every straight line perpendicular to the directrix meets the parabola, and every diameter falls wholly within it.

Let DG be perpendicular to the directrix, at any point of it, then shall DG meet the curve (figure to next proposition).

For, DF being joined, let the angle DFG be made equal to GDF, and let FG meet DG (Pl. Ge. I. 29, Cor.), which is parallel to FC, in G. The triangle DGF, having the angles at D and F equal, will also have the sides GD, GF, equal; and therefore G is a point in the curve (I. 1, Cor. 1). Again, the diameter GE falls wholly within the curve (figure to last proposition).

For, if any point O be assumed in it, it is evident that OD is greater than OF.

COR.-Hence, the two legs of the curve continually diverge from the axis.

PROPOSITION III.

A straight line bisecting the angle formed by two lines drawn from the same point in the curve, the one to the focus, and the other perpendicular to the directrix, is a tangent to the parabola in that point.

sects the vertical angle of the isosceles triangle GDF, it will also bisect the base DF at right angles (Pl. Ge. I. 4). Hence the triangles HED, HEF, are equal in every respect. Thus, HF is equal to HD, and therefore greater than HA. Consequently, the point II is without the curve (I. 1, Cor. 1). COR. 1.-Hence the method of drawing a tangent from any point in the curve.

COR. 2.-If a straight line be drawn from the focus, to any point in the directrix, the perpendicular which bisects it will touch the parabola; also, every perpendicular to it, which cuts the curve, will be nearer to the focus than to the point in the directrix.

COR. 3. The parabola is concave towards the axis. For (Cor. 2) the tangent lies between the curve and the directrix ; the curve is therefore convex towards the directrix, or concave towards the axis.

PROPOSITION IV.

Every straight line, drawn through the focus of a parabola, except the axis, meets the curve in two points.

A

D C E

B

H

F

Let FQ be a line passing through the focus, FD perpendicular to it, DA, DE, each equal to DF; AG, EH, parallel to CF, and intersected by FQ, in G, H; and let the points AF be joined. Then, because DA is equal to DF, the angles DAF, DFA, are equal; and these being taken from the right angles DAG, DFG, the remainders, the P angles GAF, GFA, are equal.

Whence the sides GA, GF, are also equal, and therefore G a point in the curve (I. 1, Cor. 1). In the same manner, may be shown that H is a point in the curve.

it

COR.-A straight line, making an indefinitely small angle with the axis, being produced, meets the curve; hence the rate of divergency must be very small.

For let MN be the given line. Through the focus draw GH parallel to it, then it will cut the curve in two points. Through one of these points, as G, draw the tangent GP,

which must be inclined to GH, and therefore to MN, and will meet MN, if produced, in some point P; and therefore MN must cut the curve.

PROPOSITION V.

If, from any point in a parabola, a straight line be drawn, not parallel to a diameter, nor bisecting the angle formed by two lines drawn from that point, the one to the focus, and the other perpendicular to the directrix, it will meet the curve in one other point, and not in more than one.

Let H be a point in the curve, and HG a line not parallel to CF, nor bisecting the angle EHF; then HG shall meet the curve in another point.

A MD E

B

For, let FL be perpendicular to GH, and it will meet AB (Pl. Ge. I. 29, Cor), since GH is not parallel to CF; also, the point D, in which it meets AB, will be different from E; since, if D, E, were the same, it would follow that HG bisects the angle EHF, contrary to the hypothesis (I. 3).

Let DA - DE, and AG parallel to EH, meet HG in G, G will be a point in the

K

L

F

H

curve. For about the centre H, with the radius HE or HF, let a circle be described intersecting FD in K, and let another circle be described through the three points A, K, F. Then, because AB touches the circle EKF in E (Pl. Ge. III. 16), the rectangle FD DK = DE2 (Pl. Ge. III. 36) = DA2; hence DA is a tangent to the circle AKF (Pl. Ge. III. 37), and therefore AG passes through its centre; but HL, which bisects the cord FK, at right angles, also passes through its centre (Pl. Ge. III. 3); consequently G is the centre of the circle AKF. Whence GA GF, and Ga point in the parabola (I. 1, Cor. 1). If GH were to meet the curve in another point, that point would be the centre of a circle passing through F, K, and touching the line AB in a point different from A or E, which is impossible; for if it touch AB in another point, as M, then FD DKDM3;

DA2; hence DM2 = DA2, or DM=DA,

but FD DK
and M must coincide with A.

COR. 1.-If a straight line be drawn through any point

within the parabola, not parallel to a diameter, it will meet the curve in two points.

For, if the line is not parallel to the directrix, it will meet it if produced, and must therefore cut the curve in one point, as H; and hence may be found the points D, E, A, and therefore also the other point of intersection, as G.

COR. 2.-At the same point, in the curve, there cannot be more than one tangent. COR. 3.-A tangent bisects the angle formed by a straight: line drawn to the focus, and another perpendicular to the directrix from the point of contact; it also bisects, and is perpendicular to the line that subtends that angle (Cor. 2, and I. 3).

COR. 4.-If a straight line from the focus be perpendi cular to a cord, it will bisect that part of the directrix which is intercepted by perpendiculars falling upon it from the extremities of the cord; and conversely. If FL be perpendicular to GH, it will bisect AE. For the circle EKF being described about the centre H, FL = LK, hence GK = GF, but GF GA; therefore G is the centre of the circle AKF, and DA2 = FD DKDE2.

PROPOSITION VI.

A straight line terminated by the parabola, and parallel to a tangent, is an ordinate to the diameter that passes through the point of contact.

fore bisects AE in D (I. 5, Cor. 4). Whence, since AG, DN, EH, are parallel, GH is bisected in N (Pl. Ge. VI. 2). COR. 1.-An ordinate to any diameter is parallel to the tangent at the vertex of that diameter. For, if GN= NH, then AD = DE; therefore DF is perpendicular to GH (I. 5, Cor. 4), and GH parallel to MQ. COR. 2. The straight line which bisects two parallel cords is a diameter.

PROPOSITION VII.

The square of a semi-ordinate, to any diameter, is equal to the rectangle under the parameter of that diameter, and the corresponding absciss.

Case 1. Let GNH be an ordinate to the axis; then GN2 =4CV.VN.

FN2 =

=

M

D

VR

For, GN2GF2. -FN2: but GF GD CN; therefore GN2 CN2 (CN+FN) (CN-FN) (Pl. Ge. II. 5, Cor.) = 2VN CF, and because 2VN: VN = 4CV: CF, 2VN.CF4CV. VN. Hence GN2-4CV.VN.

Case 2. Let GNH (figure to last proposition) be an ordinate to any other diameter MK. Then GN2=4DM MN.

G

N

H

L

For, let the tangent MQ, the perpendiculars GA, GK, HE, and the lines DH, HF, FD, be drawn. Then, because the right-angled triangles FCD, DLN, are equiangular, and MQ, LN, parallel, DF:FC=DN:DL=MN: LQ, therefore DFLQ=FC MN (Pl. Ge. VI. 16), and 2DF LQ (=2FC MN) =4CV MN. But 2DF•LQ=DH2 - HF2 (Pl. Ge. II. c.) =DE2=DA2=GK2. Consequently, GK2= 4CV MN. Again, from the similar triangles DFC, DQM, we have CF: FDDQ: DM, therefore (Pl. Ge. V. 15) CV: DQ = DQ: DM, and CV: DM = DQ2: DM2 (Pl. Ge. VI. 20, Cor. 3). But, from the similar triangles GKN, DQM, GK2: GN2DQ2: DM2 (Pl. Ge. V. 11). Whence GK2: GN2 CV: DM4CV MN: 4DM MN. But GK2=4CV. MN, consequently GN2 4DM MN.

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