§8.2#2καὶ πεποιήσθω, ὡς ἡ ΥΣ πρὸς τὴν ΣΤ, οὕτως ἡ ΑΒ πρὸς τὴν ΒΗ, πεποιήσθω δὲ καί, ὡς ἡ ΡΤ πρὸς τὴν ΤΣ, οὕτως ἡ ΑΘ πρὸς τὴν ΘΒ, καὶ γεγράφθω περὶ ἄξονα τον ΘΗ ἐπὶ μὲν τῆς πρώτης πτώσεως ἔλλειψις, ἐπὶ δὲ τῆς δευτέρας ὑπερβολή, ὥστε, οἷον ἐὰν ἐπ᾿ αὐτῆς ληφθῇ σημεῖον ὡς τὸ Γ, καὶ κάθετος ἀχθῇ ἡ Γ∠, λόγον εἶναι τοῦ ὑπὸ τῶν Θ∠Η πρὸς τὸ ἀπὸ ∠Γ τὸν συνημμένον ἔκ τε τοῦ, ὄν ἔχει ἡ ΤΣ πρὸς ΣΥ, καὶ ἐξ οὗ ὅν ἔχει ἡ ΤΣ πρὸς ΣΡ καὶ ἐξ οὗ ὄν ἔχει ὁ δοθεὶς λόγος, ὅς ἐστιν ὁ τοῦ ἀπὸ ΡΤ πρὸς τὸ ἀπὸ ΤΣ, καὶ ἤχθω ὀρθὴ ἡ ΒΚ. λέγω, ὅτι ἡ Θ Κ ποιεῖ τὸ ἐπίταγμα.
And let it be made that, as ΥΣ is to ΣΤ, so is ΑΒ to ΒΗ, and let it also be made that, as ΡΤ is to ΤΣ, so is ΑΘ to ΘΒ, and let there be described about the axis ΘΗ, in the first case an ellipse, and in the second case a hyperbola, so that, if any point is taken on it, such as Γ, and a perpendicular Γ∠ is drawn, the ratio of the rectangle under Θ, ∠, Η to the square on ∠Γ is that compounded of that which ΤΣ has to ΣΥ, and of that which ΤΣ has to ΣΡ, and of that which the given ratio has, which is that of the square on ΡΤ to the square on ΤΣ. And let the perpendicular ΒΚ be drawn. I say that ΘΚ forms the requirement.
ἤχθω γὰρ κάθετος ἡ Γ∠, καὶ πεποιήσθω, ὡς μὲν ἡ ΑΒ πρὸς τὴν ΒΗ, οὕτως ἡ ΖΒ πρὸς τὴν Β∠, ὡς δὲ ἡ ΑΘ πρὸς τὴν ΘΒ, οὕτως ἡ Ε∠ πρὸς τὴν ∠Β ὥστε ἔσται ὁ μὲν τῆς ∠Η πρὸς τὴν ΑΖ λόγος ὁ αὐτὸς τῷ τῆς ΗΒ πρὸς τὴν ΒΑ, τουτέστιν τῷ τῆς ΤΣ πρὸς ΣΥ ὁ δὲ τῆς Θ∠ πρὸς ΑΕ λόγος ὁ αὐτός ἐστιν τῷ τῆς ΤΣ πρὸς ΣΡ τὸ αὐτὸ γὰρ ἐν τῇ ἀναλύσει ἀπεδείχθη·
For let the perpendicular Γ∠ be drawn, and let it be made that, as ΑΒ is to ΒΗ, so is ΖΒ to Β∠, and as ΑΘ is to ΘΒ, so is Ε∠ to ∠Β, so that the ratio of ∠Η to ΑΖ will be the same as that of ΗΒ to ΒΑ, that is, as that of ΤΣ to ΣΥ; and the ratio of Θ∠ to ΑΕ is the same as that of ΤΣ to ΣΡ, for the same thing was proved in the analysis; so that the ratio of the rectangle under Θ, ∠, Η to the rectangle under Ζ, Α, Ε is compounded of that which ΤΣ has to ΣΥ and ΤΣ to ΣΡ.
ὥστε τοῦ ὑπὸ Θ∠Η πρὸς τὸ ὑπὸ ΖΑΕ λόγος συνῆπται ἐξ οὗ ὃν ἔχει ἡ ΤΣ πρὸς ΣΥ καὶ ἡ ΤΣ πρὸς ΣΡ. ἀλλʼ ἐπεὶ τὸ ὑπὸ Θ∠Η πρὸς τὸ ἀπὸ ∠Γ τὸν συνημμένον ἔχει λόγον ἐξ οὗ ὃν ἔχει ἡ ΤΣ πρὸς Σ καὶ ἡ ΤΣ πρὸς ΣΡ καὶ ἔτι ὁ δοθεὶς λόγος ὁ τοῦ ἀπὸ ΡΤ πρὸς τὸ ἀπὸ ΤΣ. καὶ τὸ ὑπὸ Θ∠Η πρὸς τὸ ἀπὸ ∠Γ συνῆπται ἐξ οὗ ὃν ἔχει τὸ ὑπὸ Θ∠ πρὸς τὸ ὑπὸ ΖΑΕ καὶ τὸ ὑπὸ ΖΑΕ πρὸς τὸ ἀπὸ ΔΓ, καί ἐστιν ὁ τοῦ ὑπὸ τῶν Θ∠Η πρὸς τὸ ὑπὸ ΖΑΕ λόγος ὁ αὐτὸς τῷ συνημμένῳ ἐξ οὗ ὃν ἔχει ἡ ΤΣ πρὸς ΣΥ καὶ ἡ ΤΣ πρὸς ΣΡ λοιπὸς ἄρα τοῦ ὑπὸ ΖΑΕ πρὸς τὸ ἀπὸ ∠Γ λόγος ὁ αὐτός ἐστιν τῷ τοῦ ἀπὸ ΡΤ πρὸς τὸ ἀπὸ ΤΣ, τουτέστι τῷ τοῦ ἀπὸ Ε∠ πρὸς τὸ ἀπὸ ∠Β. καὶ πάντα πρὸς πάντα·
But since the rectangle under Θ, ∠, Η has to the square on ∠Γ the ratio compounded of that which ΤΣ has to Σ[Υ] and ΤΣ to ΣΡ, and further the given ratio, which is that of the square on ΡΤ to the square on ΤΣ. And the ratio of the rectangle under Θ, ∠, Η to the square on ∠Γ is compounded of that which the rectangle under Θ, ∠[, Η] has to the rectangle under Ζ, Α, Ε and that which the rectangle under Ζ, Α, Ε has to the square on ΔΓ, and the ratio of the rectangle under Θ, ∠, Η to the rectangle under Ζ, Α, Ε is the same as that compounded of that which ΤΣ has to ΣΥ and ΤΣ to ΣΡ; therefore, the remaining ratio of the rectangle under Ζ, Α, Ε to the square on ∠Γ is the same as that of the square on ΡΤ to the square on ΤΣ, that is, as that of the square on Ε∠ to the square on ∠Β.
ὡς ἄρα τὸ ἀπὸ Α∠ πρὸς τὰ ἀπὸ Γ∠, ∠Β, οὕτως ἐστὶν τὸ ἀπὸ ΡΤ πρὸς τὸ ἀπὸ ΤΣ, τουτέστιν ὁ δοθεὶς λόγος· ὥστε τὸ ΘΚ μέρος τῆς τομῆς ποιεῖ τὸν τόπον.
And all to all; as therefore the square on Α∠ is to the [sum of the] squares on Γ∠, ∠Β, so is the square on ΡΤ to the square on ΤΣ, that is, the given ratio; so that ΘΚ, as a part of the section, forms the locus.
Τούτων οὕτως ἐχόντων ἐλευσόμεθα ἐπὶ τὸ ἐξ ἀρχῆς.
These things being so, we shall proceed to the original [problem].
ἔστω θέσει εὐθεῖα ἡ ΑΒ καὶ δοθὲν τὸ Γ ἐν τῷ αὐτῷ ἐπιπέδῳ, καὶ διήχθω ἡ ∠Γ καὶ κάθετος ἡ ∠Ε, λόγος δὲ ἔστω τῆς Γ∠ πρὸς ∠Ε. λέγω, ὅτι τὸ ∠ ἅπτεται κώνου τομῆς, καὶ ἐὰν μὲν ὁ λόγος ᾖ ἴσος πρὸς ἴσον, παραβολῆς, ἐὰν δὲ ἐλάσσων πρὸς μείζονα, ἐλλείψεως,\ ἐὰν δὲ μείζων πρὸς ἐλάσσονα, ὑπερβολῆς.
Let AB be a straight line given in position, and Γ a given point in the same plane, and let ∠Γ be drawn through, and the perpendicular ∠Ε, and let there be a ratio of Γ∠ to ∠Ε. I say that ∠ lies on a conic section, and if indeed the ratio is of equal to equal, on a parabola, and if of lesser to greater, on an ellipse, and if of greater to lesser, on a hyperbola.
ἔστω πρότερον ὁ λόγος ἴσος πρὸς ἴσον, τουτέστιν ἔστω πρότερον ἴση ἡ Γ∠ τῇ ∠Ε. δεῖξαι, ὅτι τὸ ∠ ἅπτεται παραβολῆς.
For let the ratio first be of equal to equal, that is, let Γ∠ first be equal to ∠Ε. It is required to show that ∠ lies on a parabola.
ἤχθω κάθετος ἡ ΓΖ θέσει ἄρα ἐστί· τῇ δὲ ΑΒ παράλληλος ἡ ∠Η. καὶ ἐπεὶ τὸ ἀπὸ Ε∠ ἴσον τῷ ἀπὸ ΔΓ. ἴση δὲ ἡ μὲν Ε∠ τῇ ΖΗ, το δὲ ἀπὸ ∠Γ ἴσον τοῖς ἀπὸ ∠Η, ΗΓ, τὸ ἄρα ἀπὸ ΖΗ ἴσον ἐστὶ τοῖς ἀπὸ ∠Η, ΗΓ. καὶ ἔστιν θέσει ἡ ΖΓ, καὶ δύο δοθέντα τὰ Ζ, Γ· τὸ ∠ ἄρα ἅπτεται παραβολῆς· τοῦτο γὰρ προδέδεικται.
Let the perpendicular ΓΖ be drawn; therefore it is given in position; and let ∠Η be drawn parallel to ΑΒ. And since the square on Ε∠ is equal to the square on ΔΓ. And Ε∠ is equal to ΖΗ, and the square on ∠Γ is equal to the [sum of the] squares on ∠Η, ΗΓ, therefore the square on ΖΗ is equal to the [sum of the] squares on ∠Η, ΗΓ. And ΖΓ is given in position, and Ζ, Γ are two given [points]; therefore ∠ lies on a parabola; for this has been shown beforehand.
συντεθήσεται δὴ οὕτως· ἔστω ἡ τῇ θέσει ἡ ΑΒ, τὸ δὲ δοθὲν τὸ Γ, καὶ ἤχθω κάθετος ἡ ΓΖ, καὶ θέσει οὔσης τῆς ΓΖ καὶ δύο δοθέντων των τῶν Ζ, Γ εὑρήσθω παραβολὴ ἡ ΔΘ, ὥστε, οἷον ἐὰν ληφθῇ σημεῖον ὡς τὸ ∠, ἀχθῇ δὲ κάθετος ἡ ∠Η, ἴσον ἐστὶν τὸ ἀπὸ ΖΗ τοῖς ἀπὸ ∠Η, ΗΓ. λέγω, ὅτι ἡ ∠Θ γραμμὴ τὸν τόπον ποιεῖ, τουτέστιν ὅτι, οἵα τις ἐὰν διαχθῇ ὡς ἡ Γ∠ καὶ κάθετος ἡ ∠Ε, ἴση ἐστὶν ἡ Γ∠ τῇ ∠Ε.
ἤχθω κάθετος ἡ ∠Η·
It will indeed be synthesized thus: Let AB be given in position, and Γ the given [point], and let the perpendicular ΓΖ be drawn, and ΖΓ being given in position, and Ζ, Γ being two given [points], let there be found a parabola ΔΘ, so that, if any point is taken, such as ∠, and the perpendicular ∠Η is drawn, the square on ΖΗ is equal to the [sum of the] squares on ∠Η, ΗΓ. I say that the line ∠Θ forms the locus, that is, that whatever line is drawn, such as Γ∠, and the perpendicular ∠Ε, Γ∠ is equal to ∠Ε.
διὰ ἄρα τῆς παραβολῆς ἴσον ἐστὶν τὸ ἀπὸ ΖΗ τοῖς ἀπὸ ∠Η, ΗΓ. καί ἐστιν τῇ μὲν ΖΗ ἴση ἡ Ε∠, τοῖς δὲ ἀπὸ ∠Η, ΗΓ ἴσον τὸ ἀπὸ ∠Γ· τὸ ἄρα ἀπὸ ∠Γ ἴσον ἐστὶν τῷ ἀπὸ ∠Ε. ἴση ἄρα ἐστὶν ἡ Γ∠ τῇ ∠Ε· ἡ ἄρα ∠Θ γραμμὴ ποιεῖ τὸν τόπον.
For let the perpendicular ∠Η be drawn; therefore by the parabola, the square on ΖΗ is equal to the [sum of the] squares on ∠Η, ΗΓ. And Ε∠ is equal to ΖΗ, and the square on ∠Γ is equal to the [sum of the] squares on ∠Η, ΗΓ; therefore the square on ∠Γ is equal to the square on ∠Ε. Therefore Γ∠ is equal to ∠Ε; therefore the line ∠Θ forms the locus.