Since PgST is a parallelogram, Pg=ST=SP. Now (22) cg2 = 4SP x Pg=4SP × SP=4SP. = (26.) COR. Hence QV (4SP × PV =) bc × PV; and since bc is constant with respect to the same diameter, PVQV'. That the square of the ordinate is equal to the parameter x abscissa, is therefore a general property of the Parabola. PROPERTY 9. PROPERTY 9. (27.) Draw QR parallel to PV, and meeting PT in R; then QROPR. In this case (since QV is parallel to RP) PVQR is a parallelogram; :. PR=QV, and QR=PV; but (26) PV∞QV', :. QR∞PR2. COR. From hence it follows, that if PS, PW, be two lines meeting in a PROPERTY 10. (28.) If QY be a tangent at Q, and VP be produced to meet it int, then Vt is bisected in P. (Fig. in page 19.) Produce QV to q, and draw q Y parallel to Vt; then, by sim. triangles, (since Qq=2QV) QY will be double of Qt, and q Y double of Vt. By Art. 27, Pt : qY :: Qt: QY' :: 1 : 4; But Vt=qY,.. Pt=Vt, or Vt is bisected in P. PROPERTY 11. (29.) Let fall SY perpendicular upon PT, and join then AY; AY is at right angles to AZ. Since (17) ST=SP, and SY is perpendicular to PT, it will divide the triangle PST into two equal triangles; consequently TY=YP; but (18) TA is also equal to AN; AN; . TY: YP: TA: AN; hence (Euc. 6. 2.) AY is parallel to PN, and consequently perpendicular to the line AZ. (30.) COR. Since the normal PO is perpendicular to PT, it is parallel to SY, .. TS: SO TY: YP; but TY=YP, .. TS=SO. Hence, (since TS=SP= SO) if a circle be described with center S at the distance SP, it will pass through the points P, T, and O; and the LOSP at the center will be double of the angle OTP at the circumference. PROPERTY 12. (31.) SY is a mean proportional between SA and SP. Since the angle SYT (Fig. p. 20) is a right angle, and YA perpendicular to ST, the triangle SYA is similar to the triangle SYT (Euc. 6. 8.); hence SA SY:: SY: ST (=SP). (32.) COR. Hence SY' = SA × SP, and SY= SAXSP; also 4SY'=4SA× SP=BC×SP; and as BC is constant, SY'∞ SP, and SY∞ √SP. CHAP. |