IN the Ellipsis (KDEK, Fig. 120.) the(α) 1.1 square of the semiordinate (IK) is equal to the rectangle (IL) of the Latus Rectum (EL) and the abscissa (EI) (less or) taking first out another rectangle (LS) of the same abscissa (EI or LR) and RS a fourth proportional to (DE) the Latus Transversum (EL) the Latus Rectum and (EI) the abscissa.
Suppose the side of the cone to be AB here also = a and BM parallel to the section = b and the intercepted AM=c, and EI=eb; and NI will be again = ec, all as in the hyperbola. And mak••ng also here as in the hyper∣bola MC=d, and the Latus Transversum DE=ob, so that DI will be ob−eb; then will (by reason of the simili∣tude of the ▵ ▵ BMC, DEP and DIO) EP be=od, and IO=od−ed. Therefore ▭ of NIO will be = oecd−eecd=□IK. But this square divided by the abscissa EI=eb gives 〈 math 〉〈 math 〉 or 〈 math 〉〈 math 〉 for that line IS which with the abscissa would make the rectangle ES= to the said square IK. Now therefore if we call the Latus Rectum a right line found after the same way as in the parabola, by making ac∣cording to Cons. 1. Prop. 1.