Ex. 2. Given five points on a conic, to draw the tangent at any one of them. The point F must then coincide with A, and the line QF drawn through E must therefore take the position qA. The tangent therefore must be pA.

Ex. 3. Investigate by trilinear coordinates (Art. 62) Mac Laurin's method of generating conics. In other words, find the locus of the vertex of a triangle whose sides pass through fixed points and base angles move on fixed lines. Let a, B, Y

be the sides of the triangle formed by the fixed points, and let the fixed lines be la + mẞ + ny = 0, l'a + m'ß + n'y = 0. Let the base be a μß. Then the line joining to ẞy, the intersection of the base with the first fixed line, is

And the line joining to ay, the intersection of the base with the second line, is

μ

(l'μ + m') a + n'μy = 0.

Eliminating from the last two equations, the equation of the locus is found to be Im'aß = (mẞ+ny) (l'a + n'y),

a conic passing through the points

By, ya, (a, la + mß + ny), (ß, l'a + m'ß + n'y).

EQUATION REFERRED TO TWO TANGENTS AND THEIR CHORD.

270. It much facilitates computation (Art. 229) when the position of a point on a curve can be expressed by a single variable; and this we are able to do in the case of two of the principal forms of equations of conics already given. First, let I, M be any two tangents and R their chord of contact. Then the equation of the conic (Art. 252) is LM = R2; and if μL = R be the equation of the line joining LR to any point on the curve (which we shall call the point μ), then substituting in the equation of the curve, we get M=μR and LM for the equations of the lines joining the same point to MR and to LM. Any two of these three equations therefore will determine a point on the conic.

The equation of the chord joining two points on the curve H, H', is

μμ'I − (μ + μ') R + M = 0.

For it is satisfied by either of the suppositions

(μL=R, μR=M), (μ'L=R, μ'R=M).

If μ and coincide we get the equation of the tangent, viz.

μ*L-2μR+M=0.

Conversely, if the equation of a right line (μL-2μR+M=0) involve an indeterminate u in the second degree, the line will always touch the conic LM-Ro.

271. To find the equation of the polar of any point.

The coordinates L', M', R' of the point substituted in the equation of either tangent through it give the result

Therefore the coordinates of the point of contact satisfy the

equation

ML-2RR+LM' = 0,

which is that of the polar required.

If the point had been given as the intersection of the lines aL=R, bR=M, it is found by the same method that the equation of the polar is

abL-2aR+M=0.

272. In applying these equations to examples it is useful to take notice that, if we eliminate R between the equations of two tangents

=

we get μμ'L= M for the equation of the line joining LM to the intersection of these tangents. Hence, if we are given the product of two μ's, pu' a, the intersection of the corresponding tangents lies on the fixed line aL = M. In the same case, substituting a for up' in the equation of the chord joining the points, we see that that chord passes through the fixed point (aL + M, R).

Again, since the equation of the line joining any point μ to LM is μ3L = M, the points +μ, -μ lie on a right line passing through LM.

Lastly, if LM = R, LM-R be the equations of two conics having L, M for common tangents, then since the equation μL = M does not involve R or R', the line joining the point +μ on one conic to either of the points +μ on the other, passes through LM the intersection of common tangents. We shall say that the point + on the one conic corresponds directly to the point +μ and inversely to the point on the other. And we shall say that the chord joining any two points on one conic corresponds to the chord joining the corresponding points on the other.

KK.

Ex. 1. Corresponding chords of two conics intersect on one of the chords of intersection of the conics.

The conics LM - K2, LM - R2 have R2 R2 for a pair of common chords. But the chords

μμ'L − (μ + μ') R + M = 0, μμ'L − (μ + μ') R' + M = 0, evidently intersect on R-R'. And if we change the signs of μ, ' in the second equation, they intersect on R + R'.

Ex. 2. A triangle is circumscribed to a given conic; two of its vertices move ou fixed right lines; to find the locus of the third.

Let us take for lines of reference the two tangents through the intersection of the fixed lines, and their chord of contact. Let the equations of the fixed lines be

while that of the conic is

aL-M=0, bL - M = 0,

LMR2 0.

Now we proved (Art. 272) that two tangents which meet on aL - M must have the product of their 'sa; hence, if one side of the triangle touch at the point μ, the others will touch at the points a b

and their equations will be

μ μ

μ can easily be eliminated from the last two equations, and the locus of the vertex is found to be

the equation of a conic having double contact with the given one along the line R*.

Ex. 3. To find the envelope of the base of a triangle, inscribed in a conic, and whose two sides pass through fixed points.

Take the line joining the fixed points for R, let the equation of the conic be LM = R2, and those of the lines joining the fixed points to LM be

Now, it was proved (Art. 272) that the extremities of any chord passing through (aLM, R) must have the product of their u's = a. Hence, if the vertex be М the base angles must be and and the equation of the base must be

a μ

b μ

The base must, therefore (Art. 270), always touch the conic

a conic having double contact with the given one along the line joining the given points.

Ex. 4. To inscribe in a conic section a triangle whose sides pass through three given points.

Two of the points being assumed as in the last Example, we saw that the equation of the base must be

abL (a+b) μR + μ2M = 0.

*This reasoning holds even when the point LM is within the conic, and therefore the tangents L, M imaginary. But it may also be proved by the methods of the section, that when the equation of the conic is L2+ M2 R2, that of the locus form L2+ M2 = k2R2,

Now, if this line pass through the point cL R = 0, dR — M = 0, we must have ab (a + b) μc + μ2cd = 0,

an equation sufficient to determine μ.

Now, at the point μ we have μL = R, μ2L = M; hence the coordinates of this point must satisfy the equation

abL (a + b) cR + cdM = 0.

The question, therefore, admits of two solutions, for either of the points in which this line meets the curve may be taken for the vertex of the required triangle. The geometric construction of this line is given Art. 297, Ex. 7.

Ex. 5. The base of a triangle touches a given conic, its extremities move on two fixed tangents to the conic, and the other two sides of the triangle pass through fixed points; find the locus of the vertex.

=

Let the fixed tangents be L, M, and the equation of the conic LM R2. Then the point of intersection of the line L with any tangent (u2L – 2μR + M) will have its coordinates L, R, M respectively proportional to 0, 1, 2u. And (by Art. 65) the equation of the line joining this point to any fixed point L'R'M' will be

LM' - L'M = 2μ (LR′ — L'R).

Similarly, the equation of the line joining the fixed point L"R"M" to the point (2, μ, 0), which is the intersection of the line M with the same tangent, is

2 (RM" — R"M) = μ (LM" — L′′M).

Eliminating μ, the locus of the vertex is found to be

(LM' — L'M) (LM" — L"M) = 4 (LR′ – L'R) (RM" — R′′M),

the equation of a conic through the two given points.

273. The chord joining the points μ tano, μ cot

(where

is any constant angle) will always touch a conic having double contact with the given one. For (Art. 270) the equation of the chord is

2

μL-μR (tano+cot p) + M=0,

=

which, since tan + coto 2 cosec 24, is the equation of a tangent to LM sin 24 = R at the point μ on that conic. It can be proved, in like manner, that the locus of the intersection of tangents at the points μ tano, μ cot is the conic LM = R* sin224.

Ex. If in Ex. 5, Art. 272, the extremities of the base lie on any conic having double contact with the given conic, and passing through the given points, find the locus of the vertex.

Let the conics be

LM-R20, LM sin224 - R2 = 0,

then, if any line touch the latter at the point μ, it will meet the former in the points μ tan and μ cot; and if the fixed points are μ', μ", the equations of the sides are

up' tanpL-(u' + μ tan p) R + M = 0,

μμ" cot L-(u” + μ cot p) R + M = 0.

Eliminating

με

the locus is found to be

(M — μʼR) (μ′′L – R) = tan2 6 (M — μ′′R) (μ'L — R).

274. Given four points of a conic, the anharmonic ratio of the pencil joining them to any fifth point is constant (Art. 259).

The lines joining four points ', μ", "", """ to any fiftli point μ, are

μ' (μL − R) + (M−μR)=0, μ” (μL-R)+(M — μR) = 0, μ'” (μL – R) + (M−μR)=0, μ'''' (μL – R) + (M − μR) = 0, and their anharmonic ratio is (Art. 58)

and is, therefore, independent of the position of the point μ.

We shall, for brevity, use the expression, "the anharmonic ratio of four points of a conic," when we mean the anharmonic ratio of a pencil joining those points to any fifth point on the

curve.

275. Four fixed tangents cut any fifth in points whose anharmonic ratio is constant.

Let the fixed tangents be those at the points p', μ", μ"", μ'"""', and the variable tangent that at the point u; then the anharmonic ratio in question is the same as that of the pencil joining the four points of intersection to the point LM. But (Art. 272) the equations of the joining lines are

μ'μL-M=0, μ”μL-M=0, μ""μL-M=0, μ"""μL - M=0, a system (Art. 59) homographic with that found in the last Article, and whose anharmonic ratio is therefore the same. Thus, then, the anharmonie ratio of four tangents is the same as that of their points of contact.

276. The expression given (Art. 274) for the anharmonic ratio of four points on a conic μ', p", "", """ remains unchanged if we alter the sign of each of these quantities; hence (Art. 272) if we draw four lines through any point LM, the anharmonic ratio of four of the points (μ, μ", "", ""'') where these lines meet the conic, is equal to the anharmonic ratio of the other four points (— μ', - μ", - μ"", - p'''') where these lines meet the conic.

For the same reason, the anharmonic ratio of four points on one conic is equal to that of the four corresponding points on another; nce corresponding points have the same μ (Art. 272). Again,

pression (Art. 274) remains unaltered, if we multiply each