describe a circle. Draw TL touching the circle, and let SL

meet the conic in P. Then, SLT being a right angle, TP touches the conic.

[Prop. III., Cor. Similarly, if TM be the other tangent from T to the circle, and SM meet the conic in Q, then TQ will be the tangent at Q.

PROP. VI. The tangents at P, Q intersect in T. To prove that TP, TQ subtend equal angles at S.

Let TL, TM, TN be the perpendiculars from T on SP, SQ, and the directrix.

since Tlies on the tangent at Q.

Hence, in the right-angled triangles SLT, SMT, the sides SL, SM are equal. But the hypotenuse ST is common.

Therefore the angles TSL, TSM are equal, or TP, TQ subtend equal angles at S.

Note. If the points of contact P, Q lie on opposite branches of a hyperbola, so that SL produced backwards

T

M

P

passes through P, then TSL is the supplement of the angle which TP subtends at S. In this case the tangents TP, TQ subtend supplementary angles at S.

PROP. VII. The chords PR, QR meet the directrix in p, q, and the tangents at P, Q meet the tangent at R in p', q. To prove that

LpSq=&PSQ=p' Sq'.

Since P, R are points on the same conic,

By subtraction 4 pSq = PSQ.

Again, p'R, p'P subtend equal angles at S.

By subtraction ≤p'Sq' =&PSQ=pSq from above.

[Prop. VI.

PROP. VIII. The chord of contact of tangents through T meets the directrix in R. 1o prove that TSR is a right angle.

Since P, Q, the points of contact, lie on the same conic,

COR. If a chord PQ, being produced, pass through a fixed point R on the directrix, the tangents at P, Q will meet on the fixed straight line ST, drawn at right angles to RS.

PROP. IX. To prove that SGP, SPM are similar triangles, and that

SG: SP=SA : AX,

PG being normal at P, and PM perpendicular to the directrix. Let the tangent at P meet the directrix in R. Then the circle described on PR as

Hence the triangles SPG, SMP are similar, and

PROP. X. If GK be the perpendicular upon SP from the foot of the normal at P, then PK will be equal to half the latus rectum.

Let PN meet the axis at right angles in N. Then the

right-angled triangles SKG, SNP have the angle at S

PROP. XI. If PQ be any focal chord, then

2SP.SQ=SE.PQ,

SE being half the latus rectum.

Let the normals at P, Q meet the axis in the points G, F,

those points to PQ.

and let K, M be the feet of the perpendiculars drawn from

Then FM is parallel to GK.