Similarly, if HX be made parallel to IP,

(1);

the line traversing d and b passes also through g and h, and the lines BD and bd meet in a point z on the axis major. Now

Ah2 n2AH=hd.hb

(2).

10. Given three points and two tangents, to construct the conic.

CASE I. When the given tangents traverse two of the given points. It may be deduced from Pascal's theorem that in every quadrangle inscribed within a conic the intersections of the pairs of opposite sides and of the pairs of tangents. drawn past the opposite apices or points meet in one and the same line. Thus, given the three points b, c, d (Fig. 26), and the two tangents bh and dh to describe the conic.

According to the principle just enunciated the intersection of the tangents at the opposite points b and d and of the opposite sides bc and da will meet upon one and the same line as the intersection of the opposite sides ba and cd. Wherefore, in order to determine a, produce the tangents at b and d to meet in h; draw any line hg through h, meeting the given lines bc and cd produced in ƒ and g; then the intersection a of the lines gb and fd is a fourth point on the conic. The second point a,, or the fifth point on the conic, is determined in a similar manner; so that, having found five points on the conic, we can apply the method of Art. 7 or continue the process described in this

article, using has the centre of perspective for the punctuated lines be and cd produced, d as the centre of

projection for line bc, and b as the similar centre for line cd.

CASE II. When the given points lie outside of the given tangents. Thus, given the three points B, D, C and the two tangents AG and PG to construct the curve. Draw BD (Fig. 27) cutting the tangents in H and K; CD intersecting the tangents in I and L. Divide HK at R so that

Join RS, intersecting the tangents in A and P, the points

circle, etc. Then in Fig. 28, which is a reproduction of

all that relates to the auxiliary circle in Fig. 27, we

have

kd. kb-kp2; hb.hd=ha2,

kd. kb _kp2

hb. hd ha2

Let pa or pr meet the line through h parallel to pk in z;

the lines HK and hk meet in a point on the major axis, and

therefore, upon substituting capital for small letters, we obtain

A similar proof holds for the division of IL externally in S.

COR. If the line IZ be drawn parallel to PG, it will meet the latter in a point Z at an infinite distance; so that

∞

IX.IY

IZ2= IX. IY,

which determines Z, a point on the line of contact. 11. Given five tangents, to construct the conic.

SOLUTION I.—Choose any poles S and S' on one of the tangents BB', Fig. 29, and take the two tangents BE and