ON THE SYMMETRICAL FORM OF THE EQUATION TO THE PARABOLA.*

WHEN the parabola is referred to a diameter and the tangent at its vertex, although the equation then assumes the simplest form, yet as these lines are not symmetrical with respect to the curve, the equation itself is not symmetrical with respect to the variables. In order, therefore, to get the equation under a symmetrical form, we must refer the curve to lines similarly situated with respect to it: such are two tangents to the parabola. If we take them as axes, and their intersection as origin, the equation to the curve assumes a form which bears a curious analogy to the symmetrical equations of the other conic sections and of the straight line, and is sufficiently remarkable in itself to deserve attention.

The general equation to a curve of the second degree is Ay2+ Bxy + Cx2 + Dy + Ex + F= 0 ................. (1);

the condition that this should represent a parabola is

B2 = 4AC or B≈±2 √(AC) ........

so that (1) is reduced to

(2),

{A1y ± C1x}2 + Dy + Ex+F=0............(3).

Now, let the parabola be referred to the two tangents AB, AC (fig. 7) as axes, and let AB=a, AC=b, AB being the axis of x, AC of y. Then, since AB is a tangent at B,

* Cambridge Mathematical Journal, Vol. II., p. 14.

if we make y=0 in equation (3), the two corresponding In this case equa

values of a must be each equal to ~.

tion (3) becomes

Cx2 + Ex+F=0

and the condition for its being a complete square is

E2=4CF...

and as each root is equal to a, we have

(4);

·(5);

If we take the superior sign in the first term, this equation is equivalent to

which is the equation to a straight line, or rather to two which coincide; we must therefore take the inferior sign, in order that the equation may represent a parabola. If, to both sides, the equation becomes

4xy

ab

now, we add 4

the first side of which is a complete square.

then, the square root on both sides, we have

Extracting,

the first side of which is also a complete square. Extracting the root again, we finally obtain

The form of this equation shows at once, that the curve lies wholly between the positive axes, as neither x nor y can ever become negative. So long as x <a and y<b, the positive signs only on both sides must be taken, as the difference between two fractions can never be unity. If x>a and y>b, the negative sign only on the left-hand side must be taken, as the sum of two quantities greater than unity can never be equal to unity; and either sign on the right-hand side, according to the relative magnitude of the terms on the left-hand side. If y>b and x <a, the negative sign on the first side and the positive on the second are to be taken; and if x>a and y<b, the negative sign on both sides. This apparent discontinuity, which renders it necessary to take sometimes one sign and sometimes another, arises from the equation (8) not being the complete form of the equation to the curve. All the cases are included

If we transpose one term of (8) and square both sides, we have

is the equation to a diameter passing through B, and

This form of the equation affords an easy proof of a problem in the Senate-House Papers for 1833. The enunciation is as follows: If there are three tangents to a parabola, the triangle formed by their intersection is half of that whose angular points are the points of contact.

Let ARS, BPC (fig. 7) be the triangles; then, taking the equation to the parabola referred to AB, AC as axes, the equation to the tangent is

where x,y, are the coordinates of the point P.

x=

In this equation, making successively x = 0, y = 0, we find AS=√(by), AR=√(ax1).

Now

area ASR = AR. AS sin A = √(abx ̧y1) sin C,

and

Hence

ACB-NPC-MPB- AMPN.

ACB= tab sin C,

NPC=1NC.PN sin C = x, (b−y1) sin C,

MPB={MB.PM sin C=y, (a - x1) sin C,
AMPN=x,y, sin C.

area CPB = 1⁄2 sin C' {ab — x, (b −y,) — y, (α — x,) — 2x1y1}

sin C (ab - bx, - ay,).

But, since x,y, are coordinates of a point in the parabola,

1

and multiplying by ab, and transposing,

so that

2 √(abx,y1) = ab — bx ̧ — ay1;

area CPB = sin C.2 √(abx,y,) = sin C √(abx,y,),

and therefore ASR=CPB.

Since AR=√(ax,) and AS=√(by,), we have, making

x' and y' are coordinates of the line BC; so that if from any point Q in BC we draw QS, QR parallel to the axes, the line joining the points where they cut the axes will be a tangent to the parabola. This gives the means of describing a parabola by the ultimate intersection of a line subject to move under a certain condition. For if

X Y
+ = 1

M n

be the equation to RS, m and n are subject to the condition