Again, in the two triangles DCB, DCF, because BC is equal to CF, the side DC is common to both triangles, and the angle DCB is equal to the angle DCF; therefore DB is equal to DF. But DF is equal to DE (Def. 1); hence DB is equal to DE, which is impossible (Prop. XVII., B. I.). Therefore the line AC does not meet the curve in D; and in the same manner it may be proved that it does not meet the curve in any other point than A; consequently it is a tangent to the parabola. Therefore, a tangent, &c.

Cor. 1. Since the angle FAB continually increases as the point A moves toward V, and at V becomes equal to two right angles, the tangent at the principal vertex is perpendicular to the axis. The tangent at the vertex V is called the vertical tangent.

Cor. 2. Since an ordinate to any diameter is parallel to the tangent at its vertex, an ordinate to the axis is perpen dicular to the axis.

PROPOSITION III. THEOREM.

The latus rectum is equal to four times the distance from the focus to the vertex.

Let AVB be a parabola, of which F is the focus, and V the principal_vertex; then the latus rectum AFB will be equal to four times FV.

F

Let CD be the directrix, and let AC be drawn perpendicular to it; then, according D to Def. 1, AF is equal to AC or DF, because ACDF is a parallelogram. But DV is equal to VF; that is, DF is equal to twice VF. Hence AF is equal to twice VF. same manner it may be proved that BF is equal to twice VF; consequently AB is equal to four times VF. Therefore, the latus rectum, &c.

In the

B

If a tangent to the parabola cut the axis produced, the points of contact and of intersection are equally distant from the focus.

Let AB be a tangent to the parabola GAH at the point A, and let it cut the axis produced in B; also, let AF be drawn to the focus; then will the line AF be equal tc BF.

Draw AC perpendicular to the directrix; then, since AC is parallel to BF, the angle BAC is equal to ABF. But the angle BAC is equal to BAF (Prop. II.); hence the angle ABF is equal to BAF, and, consequently, AF is equal to BF. Therefore, if a tangent, &c.

H

Cor. 1. Let the normal AD be drawn. Then, because BAD is a right angle, it is equal to the sum of the two angles ABD, ADB, or to the sum of the two angles BAF, ADB. Take away the common angle BAF, and we have the angle DAF equal to ADF. Hence the line AF is equal to FD. There fore, if a circle be described with the center F, and radius FA, it will pass through the three points B, A, D.

Cor. 2. The normal bisects the angle made by the diameter at the point of contact, with the line drawn from that point to the focus.

For, because BD is parallel to CE, the alternate angles ADF, DAE are equal. But the angle ADF has been proved equal to DAF; hence the angles DAF, DAE are equal to

each other.

Scholium. It is a law in Optics, that the angle made by a ray of reflected light with a perpendicular to the reflecting surface, is equal to the angle which the incident ray makes with the same perpendicular. Hence, if GAH represent a concave parabolic mirror, a ray of light falling upon it in the direction EA would be reflected to F. The same would be true of all rays parallel to the axis. Hence the point F, in which all the rays would intersect each other, is called the focus, or burning point.

PROPOSITION V. THEOREM.

The subtangent to the axis is bisected by the vertex.

Let AB be a tangent to the parabola ADV at the point A, and AC an ordinate to the axis; then wil BC be the subtangent, and it will be bisected at the vertex V.

B

For BF is equal to AF (Prop. IV.); and AF is equal to CE, which is the distance of the point A from the directrix. But CE is equal to the sum of CV and VE, or CV and VF.

EV F

C

Hence BF, or

BV+VF, is equal to CV+VF; that is, BV is equal to CV Therefore, the subtangent, &c.

Cor. 1. Hence the tangent at D, the extremity of the latus "ectum, meets the axis in E, the same point with the directrix. For, by Def. 8, EF is the subtangent corresponding to the tangent DE.

Cor. 2. Hence, if it is required to draw a tangent to the curve at a given point A, draw the ordinate AC to the axis. Make BV equal to VC; join the points B, A, and the line BA will be the tangent required.

PROPOSITION VI. THEOREM.

The subnormal is equal to half the latus rectum.

Let AB be a tangent to the parabola AV at the point A, let AC be the ordinate, and AD the normal from the point of contact; then CD is the subnormal, and is equal to half the latus rectum.

For the distance of the point A from the focus, is equal to its distance from the directrix, which is equal to VF+VC, or 2VF+FC; that is,

B

F C D

Also, CD is equal to FD-FC, which is equal to FA-FC (Prop. IV., Cor. 1). Hence CD is equal to 2VF, which is equal to half the latus rectum (Prop. III.). Therefore, the subnormal, &c.

PROPOSITION VII. THEOREM.

If a perpendicular be drawn from the focus to any tangent, the point of intersection will be in the vertical tangent.

Let AB be any tangent to the parabola AV, and FC a perpendicular let fall from the focus upon AB; join VC; then will the line VC be a tangent to the curve at the vertex V.

Draw the ordinate AD to the axis. Since FA is equal to FB (Prop. IV.), and FC is drawn perpendicular to AB. it divides the triangle AFB into

B

A

F

D

two equal parts, and, therefore, AC is equal to BC. BV is equal to VD (Prop. V.); hence

BCCA BV : VD,

Bu

and, therefore, CV is parallel to AD (Prop. XVI., B. IV.). But AD is perpendicular to the axis BD; hence CV is also per pendicular to the axis, and is a tangent to the curve at the point V (Prop. II., Cor. 1). Therefore, if a perpendicular,

&c.

Cor. 1. Because the triangles FVC, FCA are similar, we FV FC FC: FA;

have

that is, the perpendicular from the focus upon any tangent, is a mean proportional between the distances of the focus from the vertex, and from the point of contact.

Cor. 2. It is obvious that FV : FA :: FC2 : FA2.

Cor. 3. From Cor. 1, we have

FC FV XFA.

But FV remains constant for the same parabola; therefore the distance from the focus to the point of contact, varies as the square of the perpendicular upon the tangent.

PROPOSITION VIII. THEOREM.

The square of an ordinate to the axis, is equal to the product of the latus rectum by the corresponding abscissa.

Let AVC be a parabola, and A any point of the curve. From A draw the ordinate AB; then is the square of AB equal to the product of VB by the latus rectum.

For AB' is equal to AF-FB'. But AF is equal to VB+VF, and FB is equal to VB-VF.

=

Hence AB' (VB+VF)'—(VB-VF)', which, according to Prop. IX., Cor., B. IV., is equal to

or

4VBX VF,

VBX the latus rectum (Prop. III.). Therefore, the square, &c.

A

F

B

Cor. 1. Since the latus rectum is constant for the same parabola, the squares of ordinates to the axis, are to each other as their corresponding abscissas.

Cor. 2. The preceding demonstration is equally applicable to ordinates on either side of the axis; hence AB is equal to BC, and AC is called a double ordinate. The curve is symmetrical with respect to the axis, and the whole parabola is bisected by the axis.

PROPOSITION IX. THEOREM.

The square of an ordinate to any diameter, is equal to four times the product of the corresponding abscissa, by the distance from the vertex of that diameter to the focus.

Let AD be a tangent to the parabola VAM at the point A; through A draw the diameter HAC, and through H any point of the curve, as B, draw BC parallel to AD; draw also AF to the focus; GD then will the square of BC be equal to 4AF XAC.

Draw CE parallel, and EBG

K

V F

A

C

M

perpendicular to the directrix HK; and join BH, BF, HF. Also, produce CB to meet HF in L.

Because the right-angled triangles FHK, HCL are similar, and AD is parallel to CL, we have

HF: FK:: HC: HL

:: AC: DL.

HFXDL FK XAC,

=

Hence (Prop. I., B. II.),

or

But

2HFXDL=2FK × AC, or 4VFX AC.
2HFXDL=HL-LF' (Prop. X., B. IV.)

Hence CE' is equal to 4VFX AC.

Also, because the triangles BCE, AFD are similar, we have CE CB: DF: AF.

:

Therefore CE: CB2 :: DF: AF2 (Prop. X., B. II.)

:: VF AF (Prop. VII., Cor. 2)

:

:: 4VFX AC: 4AF XAC.

But the two antecedents of this proportion have been proven to be equal; hence the consequents are equal, or

BC2-4AF XAC.

Therefore, the square of an ordinate, &c.

of

Cor. In like manner it may be proved that the square CM is equal to 4AF XAC. Hence BC is equal to CM; and since the same may be proved for any ordinate, it follows that every diameter bisects its double ordinates.