Apollonius of Perga

Commentary on the text

1.  An ordinate from a point P to a line is any line segment joining it to a point Q on the line; the abscissa (from Latin abscindere, to cut off) corresponding to it is a segment OQ measured along the line from a fixed origin O.  These terms have now been appropriated in coordinate geometry: the point with coordinates (x, y) has abscissa x measured along the x-axis from the origin and ordinate y measured vertically to the x-axis.

2.  The diameter of the curve is the same as its axis of symmetry.  And like the diameter of a circle, any chord across the curve drawn perpendicular to this diameter is bisected by the diameter.

3.  Compare this to Proposition 2 in Archimedes' Quadrature of the Parabola.  It may look as though they say the same thing, but they do not.  Archimedes' Proposition 2 reverses the hypotheses and conclusion of Apollonius' Proposition 33: whereas Archimedes' hypothesis assumes that the tangent line has been drawn, Apollonius makes this the conclusion of the proposition.  In other words, Apollonius' Proposition 33 is the converse of Archimedes' Proposition 2.  Rightly, then, it is this proposition of Apollonius which solves the tangent problem for the parabola.  (On the other hand, Archimedes' Proposition 1 also leads to the solution of the tangent problem for the parabola, but in a different way; see the exercises.)  Apollonius uses the phrase "the line AC will fall outside the section" to signify that must be tangent to the curve.  For C is therefore the only point on the curve that lies on this line.

4.  You see that Apollonius is designing a proof by contradiction here.  This is the reason for the curious "bump" in the diagram; he is assuming that the line AC which we assert in the statement of the proposition to be tangent to the curve is not actually tangent.  Of course, his intention is to show that this assumption leads to a contradiction, whence the assertion is true.

5.  Because BG > FB.  It is also worth noting here that the translator of the text is being true to the original Greek when the notation for square of a length is written in words; the exponent notation does not appear in Greek mathematics!  Later you will notice that the product of two lengths is represented by identifying the rectangle formed by the two lengths rather than by using a multiplication sign of any sort.

6.  Because by similar triangles, FB : CD : : BA : AD.

7.  Proposition i.20 is where Apollonius gives the "symptom" of the parabola.  (This is essentially the same as Archimedes' Proposition 3 in Quadrature of the Parabola.)  It states that the ratio of the squares of the ordinates from the parabola (BG and CD) is the same as the ratio of the corresponding abscissas (BE and DE).  To see why this must be true, rewrite the proportion in algebraic form by treating AB as the x-axis of a coordinate system and the tangent line at E as the y-axis (it is irrelevant whether these axes are perpendicular to each other).  Then the points C and G on the curve have coordinates, say,

repsectively, so that

Then in these terms, the proportion states that

which means that for all points on the curve the ratio of the square of the y-coordinate to the x-coordinate is the same.  If we call this common value p, it follows that for any point (x, y) on the curve,

Of course, this is the familiar equation of a parabola.  In Apollonius' geometric perspective, it gives the symptom of the curve.
    Apollonius also interprets this by stating that the square on the ordinate (y) of a point on the parabola is equal to the rectangle formed by its abscissa (x) and a special fixed length (p) called the parameter of the curve.  In the language he uses, the rectangle is "applied" to the abscissa; since the Greek word for "application" is parabole, we understand where he gets the name "parabola" for the curve.

8.  Because AD is twice EA, which also equals DE.

9.  Follow the link to Elements ii.5.  In modern language, it says that if we divide a length a into pieces x and a - x, then the rectangle on the pieces, that is  x(a - x), plus the square on the segment between the division point and the midpoint of the segment, that is (x - a/2 )(x - a/2 ), equals

which is the square on the half of the segment.  By removing the squared term on the left side of the equation, we see that the rectangle on the parts,  x(a - x), is strictly less than the square on the half of the segment, .  This is what Apollonius is using here: AB is cut into unequal pieces at E, so that the rectangle on AE and BE is less that the square on half of AB; equivalently, 4 times the rectangle is less than the square on all of AB.
    The contradiction appears in reconciling the last three proportions in the text: on the one hand,

but the ratio on the right side is unity since  4 rect. DE, EA = sq.AD, while the ratio on the left is less than unity since  4 rect. BE, EA < sq. AB.

10.  The transverse side of the conic is the finite segment on the diameter between the two points where it cuts the curve.  In the diagrams in the text, the transverse side is marked AB for the hyperbola, ellipse and circle.
    In note 7, we gave the etymology of the word "parabola", which derives from the symptom of the curve.  Similarly, Apollonius gives the symptoms of the hyperbola and ellipse.  In each case, besides the length of the transverse side--call it t--there is a fixed line segment of length p which arises in his analysis which he similarly calls the parameter of the curve.  Then, for any point with abscissa x and ordinate y, the symptom of the hyperbola is equivalent to the equation

11.  Given the point C on the conic, D is the foot of the perpendicular from C to the diameter AB, but the point E has to be constructed so as to obey the proportion indicated.  How is this done?
    For the hyperbolic case, consider the configuration of the diagram below, which is identical to the figure in the text, except that the hyperbola has been omitted.  We include the point c which is diametrically across from C on the curve.  After forming triangle ACc, the line cB is drawn to cut AC in F; the foot of the perpendicular from F to AB is the desired point E.  Why?  Since triangles AEF and ACD are similar, AD : AE = CD : FE.  But CD = cD, so AD : AE = cD : FE = BD : BE because triangles BcD and BFE are similar.  This last proportion can be alternately expressed AD : BD = AE : BE, precisely the proportion Apollonius gives.

12.  We end our selection in mid-proof here, for we will not be using this result in the sequel.
 

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