"The three opposite diagonals of every hexagon circumscribing a conic intersect in a point.”

For if T be the point of contact of AB the pentagon may be considered as a hexagon AT, TB, BC, CD, DE, EA, and therefore AC, BE and DT must meet in a point ; and conversely if L is the point of intersection of AC and BE, DL must pass through T, the point of contact of AB, and similarly for the remaining sides.

PROBLEM 85. To describe an ellipse, four tangents AB, BC, CD, DA and a point E on the curve being given (Fig. 80).

[The point E must lie within the quadrilateral ABCD, which must not be a parallelogram.]

Let BE, CE meet AD in B, and C, respectively.

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Fig.80.

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Find X the centre, and P and P, the foci of the involution AC1 and DB1. Prob. 13.

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Then the tangent at E must pass through P or P, and the problem reduces to the preceding.

There are two solutions.

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In the figure B ̧c on B1B=B ̧С1; Ad on a parallel to BB is equal to AD and cd intersects AB, in X, the required centre. XP is a mean proportional between XA and XC,.

Also BP and AF intersect in L and CL will pass through T, the point of contact of AP.

138

GIVEN FIVE POINTS.

PROBLEM 86. To describe an ellipse to pass through five given points ABCDE (Fig. 81).

[No point must lie inside the quadrilateral formed by the other four.]

Let AB, DC meet in F and AC, BE in G.

Draw FG meeting DE in P. P will be a point on the tangent at A.

Similarly if BC and ED meet in H and AC, BD in K, HK will meet EA in Q, a point on the tangent at B.

The problem can evidently be completed in various ways by preceding problems.

The construction depends on Pascal's well-known theorem : "The three intersections of the opposite sides of any hexagon inscribed in a conic section are in one right line." For the tangent at A may be considered as meeting the curve in two consecutive points A and a, and therefore P, the intersection of Aɑ and DE, must lie on FG, the straight line through the intersections of AB and DC and of BE and Ca.

This line is known as the Pascal line.
There is only one solution.

PROBLEM 87. To describe an ellipse, four points on the curve A, B, C, D and a tangent ad being given (Fig. 82).

[All the points must lie on the same side of the tangent.]

Draw AB meeting ad in a, BC meeting ad in b, DC meeting it in c, and AD meeting it in d.

Find X the centre, and P and P, the foci of the involution ac and bd.

P or P, will be the point of contact of the given tangent and the problem may be completed by several preceding ones.

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In the fig. ab on aA=ab; de, on a parallel to aA=dc, and b1c, intersects ad in X, the required centre. XP=Xp, a mean proportional between Xc and Xa.

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If DC, BP meet in F, and BC, PA in G, then FG and DA will intersect in H, a point on the tangent at B.

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There are of course two solutions, as either P or P1 may be taken as the point of contact.

That P, the point of contact, is a focus of the involution is proved in Chapter 8.

POLE AND POLAR.

It has been shewn in the case of the circle (Cor. 3, p. 31) that the pairs of tangents drawn at the extremities of any chord through a fixed point intersect in a straight line.

This is also true in the case of any conic section, for let V (fig. 83) be any point in a conic and C the centre, and let CV

meet the curve in P.

Take T in CV produced such that CV CP CP: CT, and through V draw the chord QVQ, parallel to the tangent at P.

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QQ, will be the chord of contact of the pair of tangents drawn from T to the conic, and will be bisected in V.

Through V draw any chord AVB and let the tangents at A and B intersect in T.

Join CT, and draw PN parallel to AB, meeting CT, in N. Then if CT, meet AB in K and the tangent at P in L,

CK. CT-CN. CL. (Prop. 3, p. 107.)

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hence TT, is parallel to PL, and therefore T1, the intersection of the tangents at the extremities of any chord through V, lies on a fixed line.

DEF. As in the circle, the line TT, is called the polar of the point V with respect to the conic and the point V is called the pole of TT with respect to the conic.

If the pole lies without the conic (as T), its polar is the line QQ, parallel to the tangent at the point (P) where CT meets the conic, and meeting CT in a point V such that

CV: CP :: CP : CT,

i.e. is the chord of contact of tangents from the pole.

If the conic be a parabola, since the centre may be considered as at an infinite distance, the line VT must be drawn parallel to the axis meeting the curve in P and PT be made equal to PV, the polar of V will then be parallel to the tangent at P and will pass through T

If the pole be on the curve, the polar is the tangent at the point.

The directrix is the polar of the corresponding focus.

If a point (as T1) lies on the polar of V, the polar of T1 passes through V.

The following important harmonic properties should be noticed. PROP. 5. A straight line drawn through any point is divided harmonically by the point, the curve, and the polar of the point.