IN to IC, and alternatively FO to IN as FC to IC i. e. as AC to CE or AB to DE. Therefore also the square of FO (or HN) will be to the square of IN, as the square of AB to the square of DE, by Prop. 22. lib. 1. i. e. as EL the Latus Rectum to ED the Latus Transversum, by Prop. 35. lib. 1. But the □ IN is = DNE from the proporty of the circle. Therefore □ FO (or of the semiordinate HN) is to the ▭ DNE as EL the Latus Rectum to ED the Latus Transv. therefore by ver∣tue of the pres. Prop. the point H is in the Ellipsis, and so any other, &c. Q. E. D.
CONSECTARY III.
NOW if in the ellipsis the □ of AC the second Ax (= 〈 math 〉〈 math 〉 by Consect. 1.) and □ CN the distance of the Fo∣cus from the centre (= 〈 math 〉〈 math 〉 by Censect 3. Prop. 5. the figure whereof you may see n. 124.) be joined in one sum; the □ AN will be = 〈 math 〉〈 math 〉, and so the line AN=〈 math 〉〈 math 〉 i. e. to half the Latus Transversum: So that hence having the axes given you may find the Foci, if from A at the interval CD you cut the transverse ax in N and N.
CONSECTARY IV.
NOW if, on the contrary, in an hyperbola (Fig. 123.) the □ AC or EF=〈 math 〉〈 math 〉 be substracted from the □ CF or CN=〈 math 〉〈 math 〉 by vertue of Consect. 2. Prop. 5. there will remain 〈 math 〉〈 math 〉 and its root 〈 math 〉〈 math 〉, i. e. half the Latus Trans∣versum CD: So that here also, the axes being given, you may find the Focus's, if from the vertex E you make EF a per∣pendicular to the ax = to the second Ax AC, and at the in∣terval CF from the centre C you cut the Latus Transversum continued in N and N.