and or Also the triangles MPF, M'PF" are similar, SP: PM :: AS: AX :: HP : PM' ; Hence (Euclid vi. 7) SPF, HPF" are similar triangles, and the angles SPF, HPF' are equal. COR. The tangents at the extremities of the axes are at right angles to them. 8. The feet of the perpendiculars from the foci on any tangent lie on the auxiliary circle. E A Let SY the perpendicular from S on the tangent at P meet HP produced in K. Join SP, CY. Then in the right-angled triangles SYP, KYP, PY is common and the angle SPY the angle HPZ (by the last proposition) = = the vertical opposite angle KPY. Hence SY KY, and SP= KP: = :. KH=KP+PH= SP+PH=2AC. And C, Y being the middle points of SH, SK, CY is parallel to HK and half of it and therefore = AC. Hence Y lies on the circle on AA' as diameter (the auxiliary circle). And similarly Z the foot of the perpendicular from H, COR. 1. CY being parallel to HP and bisecting SH, also bisects SP. Hence SYP being a right angle, the circle on SP as diameter passes through Y, and has its centre on CY; hence it touches the auxiliary circle at Y. COR. 2. If CD be drawn parallel to the tangent at P and therefore conjugate to CP, and intersect PH in E, then CEPY is a parallelogram, and PE= CY=AC. 9. If SY, HZ are perpendiculars on the tangent at P from the foci S, H; SY. HZ= BC2. Since YZH is a right angle, YC produced will meet ZH produced on the auxiliary circle (at Y'). The triangles SCY, HCY' are equal in all respects, and SY. HZ=HY'. HZ= AH. A'H= A'C2 — CH2 = BC2. 10. SP. HP= CD2. If we draw PF in fig. §8 perpendicular to DC (produced if necessary) we shall have the triangle PEF similar to HPZ and SPY; and PE AC: hence = SP: SY:: HP: HZ :: PE:PF:: AC: PF :: CD: BC, ::: PF.CD=AC.BC; .. SP. HP: SY. HZ:: CD: BC2, and SY. HZ= BC, 11. If from any point Q of the tangent PK perpendiculars QN, QT be drawn to the directrix and SP, then ST: QN AS: AX. For QT is parallel to KS; and if we draw PM perpendicular to the directrix, ST: SP:: QK: PK:: QN: PM, .. ST: QN :: SP : PM :: AS: AX. The student of analytical geometry will see in this proposition the basis of the polar equation to the tangent in terms of the angle M N K X The proposition is due to Professor Adams, and the property is equally true of all the Conic Sections. J. C. S. 3 12. Hence if QP, QP' be the tangents to an ellipse from an exterior point Q, QP, QP subtend equal angles at S. For if we draw QT, QT, QM the perpendiculars on SP, SP and the directrix, we shall have ST, ST" in the same proportion to QM and therefore equal. Hence the right-angled triangles QST, QST" are equal in all respects, and the angles QSP, QSP' equal. M X པ T P If Qlie beyond the directrix, T, T" will lie in PS, P'S produced, and the angles QSP, QSP' will be proved equal by proving their supplements equal. 13. If QP, QP' be the tangents to the ellipse from Q, the angles SQP, HQP' are equal. Produce SP, HP' to R, R'; and let HP, SP' intersect in 0. R P Then QP, QS bisect the angles HPR, P'SP respectively, also OPR= sum of the interior opposite angles OSP, SOP, .. angle SOP=twice SQP. Similarly the angle HOP' twice HQP'. And the angle SOP=HOP', :. SQP=HQP'. 14. Any two tangents at right angles to one another intersect on a circle whose centre is C and square on the radius = CA+ CB2. Let any two tangents at right angles to one another cut R R A the auxiliary circle in Y, Z; Y, Z. Draw SY, SY'; HZ, HZ. Let QC cut the circle in R, R'. Then SQ, HQ will be rectangles, with opposite sides equal. |