Advanced Algebra and

Paul Yiu

Department of Mathematics Florida Atlantic University

Fall 2016

November 14, 2016

Contents

12 Conics 301 12.1 ...... 301 12.1.1 Chords and tangents ...... 302 12.1.2 Triangle bounded by three tangents ...... 304 12.2 ...... 306 12.2.1 The directrices and eccentricity of an ...... 306 12.2.2 The auxiliary and eccentric angle ...... 307 12.2.3 Tangents of an ellipse ...... 307 12.2.4 Ellipse inscribed in a triangle ...... 308 12.3 ...... 311 12.3.1 The directrices and eccentricity of a ...... 311 12.3.2 Tangents of a hyperbola ...... 312 12.3.3 Rectangular hyperbolas ...... 313 12.3.4 A theorem on the tangents from a point to a conic ...... 314

13 General conics 317 13.1 Classification of conics ...... 317 13.2 Pole and polar ...... 318 13.3 Condition of tangency ...... 319 13.4 Parametrized conics ...... 320 13.5 Rectangular hyperbolas and the nine-point circle ...... 321

14 Conic solution of the (O, H, I) problem 325 14.1 Ruler and compass construction of circumcircle and incircle ...... 325 14.2 The general case: intersections with a conic ...... 325 14.3 The case OI = IH ...... 328

Chapter 12

Conics

12.1 Parabolas

Given a point F (focus) and a line L (directrix), the locus of a point P which is equidistant from F and L is a P. Let the distance between F and L be 2a. Set up a Cartesian coordinate system such that F =(a, 0) and L has equation x = −a. The origin O is clearly on the locus. Its distances from F and L are both a.

L

Q P (x, y)

directrix

− a O F =(a, 0)

Let P (x, y) be a point equidistant from F =(a, 0) and L . Then (x − a)2 + y2 = (x + a)2. Rearranging terms, we have

y2 =4ax.

The parabola P can also be described parametrically: x = at2, y =2at.

We shall refer to the point P (t):=(at2, 2at) as the point on P with parameter t.Itisthe intersection of the line y =2at and the perpendicular bisector of FQ, Q =(−a, 2at) on the directrix. 302 Conics

12.1.1 Chords and tangents

The line joining the points P (t1) and P (t2) has equation

2x − (t1 + t2)y +2at1t2 =0.

Letting t1,t2 → t, we obtain the equation of the tangent at P (t):

x − ty + at2 =0. ( ) 1 (− 2 0) The tangent at P t has slope t and intersects the axis of the parabola at at , .It can be constructed as the perpendicular bisector of FQ, where Q is the pedal of P on the directrix.

Exercise

1. (Focal chord) The line joining two distinct points P (t1) and P (t2) on P passes through the focus F if and only if t1t2 +1=0.

2. (Intersection of two tangents) Show that the tangents at the points t1 and t2 on the 2 parabola y =4ax intersect at the point (at1t2,a(t1 + t2)).

3. Justify the construction suggested by the diagram below for the construction of the tangent at a point P on a parabola.

L

P

O F

4. (Tangents from a point to a parabola) Given a point P , construct the circle with diameter PF and let it intersect the tangent at the vertex at Q1 and Q2. Then the lines PQ1 and PQ2 are the tangents from P to the parabola.

5. Let P1P2 be a focal chord of a parabola P with midpoint M. The tangents at P1 and P2 intersect at Q. Show that (i) Q lies on the directrix, (ii) these tangents are perpendicular to each other, and (iii) QM is parallel to the axis of the parabola, and (iv) QM and intersects P at its own midpoint. 12.1 Parabolas 303

T1

Q1

P

O F

Q2 T2

6. Justify the following construction of the chord of a parabola which has a given point M as its midpoint: Let Q be the pedal of M on the directrix. Construct the circle M(Q) to intersect the perpendicular from M to QF . These two intersections are on the parabola, and their midpoint is M.

L P1

Q M

O F

P2

7. A circle on any focal chord of a parabola as diameter cuts the curve again at two points P and Q. Show that as the focal chord varies, the line PQ passes through a fixed point. 8. (Chords orthogonal at the vertex) Let PQ be a chord of a parabola with vertex O such that angle POQis a right angle. Find the locus of the midpoint of PQ. 9. Find the locus of the point whose two tangents to the parabola y2 =4ax make a given angle α. 10. PQ is a focal chord of a parabola with focus F . Construct the through F tangent to the parabola at P and Q. What is the locus of the second intersection of the circles (apart from F )? 304 Conics

12.1.2 Triangle bounded by three tangents

2 Proposition 12.1. The points P (t1), P (t2), P (t3), P (t4) on the parabola y =4ax are concyclic if and only if t1 + t2 + t3 + t4 =0.

Theorem 12.2. The circumcircle of the triangle bounded by three tangents to a parabola passes through the focus of the parabola.

L

P1

A3

L3 P2

A2 F

A1

L1

P3 L2

=1 2 3 = − + 2 =0 Proof. For i , , , the tangent at ti is the line Li x tiy ati . We find λ1, λ2, λ3 such that

λ1L2 · L3 + λ2L3 · L1 + λ3L1 · L2 =0 represents a circle. This requires the coefficients of x2 and y2 to be equal, and that of xy equal to 0:

(1 − t2t3)λ1 +(1− t3t1)λ2 +(1− t1t2)λ3 =0, (t2 + t3)λ1 +(t3 + t1)λ2 +(t1 + t2)λ3 =0.

Solving these equations, we have 1 − t3t1 1 − t1t2 1 − t1t2 1 − t2t3 1 − t2t3 1 − t3t1 λ1 : λ2 : λ3 = : : t3 + t1 t1 + t2 t1 + t2 t2 + t3 t2 + t3 t3 + t1 2 2 2 =(t2 − t3)(1 + t1):(t3 − t1)(1 + t2):(t1 − t2)(1 + t3).

2 2 With these values of λ1, λ2, λ3, we compute the common coefficient of x and y as 2 (t2 − t3)(1 + t1)=−(t2 − t3)(t3 − t1)(t1 − t2). 12.1 Parabolas 305

Also,

coefficient of x 2 2 2 = (t2 − t3)(1 + t1) · a(t2 + t3)

= a(t2 − t3)(t3 − t1)(t1 − t2)(1 + t2t3 + t3t1 + t1t2), coefficient of y 2 2 2 = − (t2 − t3)(1 + t1) · a(t2 + t3)

= a(t2 − t3)(t3 − t1)(t1 − t2)(t1 + t2 + t3 − t1t2t3), the constant term 2 2 2 2 = (t2 − t3)(1 + t1) · a t2t3 2 = − a (t2 − t3)(t3 − t1)(t1 − t2)(t2t3 + t3t1 + t1t2).

Cancelling a common factor −(t2 − t3)(t3 − t1)(t1 − t2), we obtain the equation of the circle as 2 2 2 x + y − a(1 + σ2)x − a(σ1 − σ3)y + a σ2 =0, where σ1, σ2, σ3 are the elementary symmetric functions of t1, t2, t3. This circle clearly passes through the focus F =(a, 0).

Exercise 1. Prove that the orthocenter of the triangle formed by three tangents to a parabola lies on the directrix.

2. Show that the parabola tangent to the internal and external bisectors of angle B and C of triangle ABC has focus at vertex A and directrix the line BC. 1

Exercise 1. (Parabolas with a common vertex and perpendicular axes) The two parabolas P : y2 =4ax and P : x2 =4by have a common vertex O and perpendicular axes. Let A be their common point other than O. The tangent at A to P intersects P at B and the tangent at A to P intersects P at C. Show that the line BC is a common tangent of the parabolas.

2. (Conformal focal parabolas) Two parabolas have a common focus F ; their directrices intersects at a point A. Show that the perpendicular bisector of AF is the common tangent of the two parabolas.

1 Solution. Let I, Ib, Ic be the incenter and the B-, C-excenters of triangle ABC. The two bisectors of angle B and the internal bisector of angle C bound the triangle IBIc. The two bisectors of angle C and the internal bisector of angle B bound the triangle ICIb. Apart from I, the circumcircles of these triangles intersect at A. This is the focus of the parabola. The reflections of A in the bisectors of angles B and C are points on the sideline BC. Therefore, the line BC is the directrix. 306 Conics

3. (a) Find the condition for the two parabolas y2 =4ax and y2 =4b(x − c) to have a common tangent. (b) Construct the common tangent of the two parabolas.

12.2 Ellipses

Given two points F and F in a , the locus of point P for which the distances PF and PF have a constant sum is an ellipse with foci F and F . Assume FF =2c and the constant sum PF+PF =2a for a>c. Set up a coordinate system such that F =(c, 0) and F =(−c, 0). A point (x, y) is on the ellipse if and only if (x − c)2 + y2 + (x + c)2 + y2 =2a.

This can be reorganized as 2

x2 y2 + =1 with b2 := a2 − c2. (12.1) a2 b2

b P a

c a 2  2  O F − a A  O F A a F c F c

12.2.1 The directrices and eccentricity of an ellipse

2 =( ) x2 + y =1 3 If P x, y is a point on the ellipse a2 b2 , its square distance from the focus F is 2 |PF| = c x − a . a c := c = · − a Writing e a ,wehavePF e x e . The ellipse can also be regarded as the locus = a of point P whose distances from F (focus) and the line x e (directrix) bear a constant 2 | | = + a =(− 0) ratio e (the eccentricity). Similarly, PF e x c . The point F ae, and the = − a line x e form another pair of focus and directrix of the same ellipse. 2 2 2 2 2 From (x + c) + y =2a − (x − c) + y ,wehave 2 2 2 2 2 2 2 (x+ c) + y =4a +(x − c) + y − 4a (x − c) + y ; 4a (x − c)2 + y2 =4a2 − 4cx; a2((x − c)2 + y2)=(a2 − cx)2; 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 (a − c )x + a y = a (a − c ). Writing b := a −c ,wehave b x + a y = a b . 2 2 2 3 2 2 2 2 2 x b 2 2 2 c 2 2 PF =(x − c) + y =(x − c) + b 2 − 1 = 1+ 2 x − 2cx + c − b = 2 x − 2cx + a = a a a c 2 a x − a . 12.2 Ellipses 307

P

2  2 − a A  O F A a c F c

12.2.2 The auxiliary circle and eccentric angle The circle with the major axis as diameter is called the auxiliary circle of the ellipse. Let P be a point on the ellipse. If the perpendicular from P to the major axis intersects the auxiliary circle at Q (on the same side of the major axis), we write Q =(a cos θ, a sin θ) and call θ the eccentric angle of P . In terms of θ, the coordinates of P are (a cos θ, b sin θ). This is a useful parametrization of the ellipse.

Q b P Q

−a θ a O −c O c

−b

12.2.3 Tangents of an ellipse Construction of tangent at P

=( ) x1x + y1y =1 If P x1,y1 , the tangent to the ellipse at P is the line a2 b2 . The corresponding 2 2 2 point Q on the auxiliary circle is Q =(x1,y2). The tangent to the circlex +y = a at Q 2 a2 x1x + y2y = a x T = , 0 is the line . Both tangents intersect the -axis at x1 . Therefore, from the tangent of the circle (which is perperpendicular to the radius OQ), we have the point T . Then, PT is the tangent to the ellipse at P .

Q b P

−a θ −c O c a T

−b 308 Conics

Remarks. (1) This construction can be reversed to locate the point of tangency on a line which is known to be tangent to an ellipse. (2) In terms of the eccentric angle of P , these tangents are cos θ · x +sinθ · y =a, (circle) cos θ sin θ · x + · y =1, (ellipse). a b

Construction of tangents from an external point

P Q

 F O F Q

1. The circles with diameters PF and PF are tangent to the auxiliary circle at two points on the tangent to the ellipse at P .

b P P

a  F O F 2  A  O F A a c F c

2. Prove the following reflection property of the ellipse.

12.2.4 Ellipse inscribed in a triangle Theorem 12.3. If a, b, c are the three complex roots of a cubic polynomial f(z), then the two roots of its derivative f (z) are the foci of the ellipse which is tangent to the sides of the triangle with vertices a, b, c at their midpoints. 12.2 Ellipses 309

α

Ω γ β

G  Ω

 γ β α

Proof. We may assume α + β + γ =0so that the centroid is G =0and the cubic has the form f(z)=z3 + pz + q. Then, its three roots are

α = v + w, β = ωv + ω2w, γ = ω2v + ωw, for two complex numbers v and w, and ω is a primitive cube root of unity. Note that

p = αβ + γα + βγ = −(v + w)2 + v2 + w2 − vw = −3vw, q = − αβγ =(v + w)(v2 + w2 − vw)=v3 + w3.

This means f(z)=z3 − 3vw · z +(v3 + w3) and f (z)=3(z2 − vw). The two roots of f (z) are therefore the square roots of vw: √ √ Ω= vw, Ω = − vw.

β+γ = − 1 ( + ) The midpoint α of the side βγ is 2 2 v w . Therefore, 1 √ 1 √ √ | − Ω| = − ( + ) − = − ( + )2 α 2 v w vw 2 v w 1 √ √ 1 √ √ √ √ = | + |2 = ( + )( + ) 2 v w 2 v w v w 1 √ √ √ √ √ √ = (| |2 + | |2 + + ) 2 v w v w v w 1 √ √ √ √ = (|v| + |w| + v w + v w); 2 1 √ | − Ω| = − ( + )+ = α 2 v w vw ... 1 √ √ √ √ = (| | + | |− − ) 2 v w v w v w . 310 Conics

From these, |α − Ω| + |α − Ω| = |v| + |w|.

Similarly, we obtain the same expression replacing α by β and γ (the midpoints of γα and αβ). Thus Ω and Ω are the foci of an ellipse through the midpoints of the triangle.

Exercise

1. An ellipse with semi-axes a and b slides on the positive x- and y-axes. Find the locus of its center.

2 E : x2 + y =1 2. (Director circle) The tangents to the ellipse a2 b2 with slope m are √ y = mx ± m2a2 + b2.

Deduce that the locus of points from which the two tangents to E are perpendicular is the circle x2 + y2 = a2 + b2.

2 = + E : x2 + y =1 3. (Pair of tangents) (a) Show that the line y mx c touches the ellipse a2 b2 if (and only if) c2 = m2a2 + b2. (b) Deduce that if (p, q) is outside the ellipse, and p2 = a2, then the two tangents from (p, q) to the ellipse are given by

(qx − py)2 = b2(x − p)2 + a2(y − q)2.

4. (Chords of ellipse subtending a right angle at the center)

2 x + y =1 E : x2 + y =1 The line p q intersects the ellipse a2 b2 at two points P and Q such ∠ = π that POQ 2 if and only if

1 1 1 1 + = + . p2 q2 a2 b2

Construct the circle, center O, tangent to a line joining two vertices of the ellipse (not on the same axis). A chord of the ellipse cut out by a tangent of this circle subtends a right angle at the center.

5. Let B and B be the endpoints of the minor axis of an ellipse. P and Q are points on the ellipse such that BP and BQ are perpendicular. Show that the lines BP and BQ intersect on a fixed line. 12.3 Hyperbolas 311

b

P c  O a2 a F F c

12.3 Hyperbolas

Given two points F and F in a plane, the locus of point P for which the distances PF and PF have a constant difference is a hyperbola with foci F and F . Assume FF =2c and the constant difference |PF − PF| =2a for a

This can be reorganized as

x2 y2 − =1 with b2 := c2 − a2. (12.2) a2 b2

12.3.1 The directrices and eccentricity of a hyperbola

2 x2 y If P =(x, y) is a point on the hyperbola 2 − 2 =1, its square distance from the focus a b 2 4 | | = c − a := c = · − a F is PF a x c . Writing e a ,wehavePF e x e . The hyperbola can = a also be regarded as the locus of point P whose distances from F (focus) and the line x e 2 1 | | = + a (directrix) bear a constant ratio e> (the eccentricity). Similarly, PF e x c . The =(− 0) = − a point F ae, and the line x e form another pair of focus and directrix of the same hyperbola. There is an easy parametrization of the hyperbola: P (θ)=(a sec θ, b tan θ). Given a point T on the auxiliary circle C (O, a), construct (i) the tangent to the circle at T to intersect the x-axis at A, (ii) the line x = b to intersect the line OT at B, (iii) the “vertical” line at A and the “horizontal” line at B to intersect at P . The point P has coordinates (a sec θ, b tan θ), and is a point on the hyperbola H . If Q is the pedal of T on the x-axis, then the line PQis tangent to the hyperbola.

2 2 2 4 2 2 2 2 2 x b 2 2 2 c 2 2 PF =(x − c) + y =(x − c) + b 1 − 2 = 1 − 2 x − 2cx + c + b = 2 x − 2cx + a = a a a c 2 a x − a . 312 Conics

B P

T

θ −c O Q a b c A

12.3.2 Tangents of a hyperbola

2 H : x2 − y =1 Let P be a point on the hyperbola a2 b2 . The circles with diameters PF1 and PF2 are tangent to the auxiliary circle C (O, a). The line joining the points of tangency is tangent to H at P .

P

 a F O F

Exercise

1. If P and Q are on one branch of a hyperbola with foci F and G, show that F and G are on one branch of a hyperbola with foci P and Q.

2. Let P be a point on a hyperbola. The intersections of the tangent at P with the asymptotes, and the intersection of the normal at P with the axes lie on a circle through the four points passes through the center of the hyperbola. a2+b2 sec a2+b2 tan The center of the circle is 2a θ, 2b θ . The locus is the hyperbola

4(a2x2 − b2y2)=(a2 + b2)2. 12.3 Hyperbolas 313

P

 a F O F

12.3.3 Rectangular hyperbolas The line joining the points t, t of the rectangular hyperbola xy = c2 is

x + tty − c(t + t)=0.

2 Example 12.1. Four points on the rectangular hyperbola xy = c with parameters t1, t2, t3, t4 form an orthocentric system if and only if t1t2t3t4 = −1.

Proof. Consider three points on the hyperbola with parameters t1, t2, t3. The perpendicu- lars from each one of them to the line joining the other two are the lines

2 t1t2t3x − t1y + c(1 − t1t2t3)= 0, 2 t1t2t3x − t2y + c(1 − t1t2t3)= 0, 2 t1t2t3x − t3y + c(1 − t1t2t3)= 0. c t4 t1t2t3t4 = −1 (x, y)= ct4, If is such that , then it is easy to see that t4 satisfies each of the above equations. Therefore, the orthocenter is the point with parameter t4 on the hyperbola.

2 Example 12.2. Four points on the rectangular hyperbola xy = c are with parameters t1, t2, t3, t4 are concyclic if and only if t1t2t3t4 =1.

Proof. If the four points are on a circle with equation

2 2 x + y +2gx +2fy + c0 =0 then t1, t2, t3, t4 are the roots of the equation

2 2 2 c 2cf c t + +2cgt + + c0 =0, t2 t or 2 4 3 2 2 c t +2cgt + c0t +2cft + c =0. = c2 =1 From this we conclude that t1t2t3t4 c2 . 314 Conics

Example 12.3. PQis a diameter of a rectangular hyperbola. The circle, center P , passing through Q, intersects the hyperbola at three points which form the vertices of an equilateral triangle.

Proof. Let P and Q be the points with parameters τ and −τ. Since the points with param- eters −τ, t1, t2, t3 are concyclic, −τ · t1t2t3 =1, and t1t2t3 · τ = −1. This means that P is the orthocenter of the triangle with vertices t1, t2, t3. This coincides with the circumcenter, and the triangle is equilateral.

Exercise

1. If P and Q are on one branch of a hyperbola with foci F and G, show that F and G are on one branch of a hyperbola with foci P and Q. c c 2. (p, q) ct1, ct2, The pedal of the point on the line joining t1 and t2 is the point 2 2 t1t2p − t1t2q +(t1 + t2)c −t1t2p + q + t1t2(t1 + t2)c 2 2 , 2 2 . 1+t1t2 1+t1t2

3. ABC is triangle with orthocenter H, inscribed in a rectangular hyperbola with asymp- totes L and L . Let X, Y , Z be the pedals of A, B, C on L . Show that the perpen- diculars from X to BC, Y to CA, and Z to AB are concurrent at the pedal of H on L .

12.3.4 A theorem on the tangents from a point to a conic

Theorem 12.4. Let P be the intersection of the tangents at T1 and T2 of an ellipse with foci F and F . The lines PF and PF make equal angles with the tangents PT1 and PT2.

P

T1

T2

F  F 12.3 Hyperbolas 315

=( ) x1x + y1y =1 Proof. (1) Let T1 x1,y1 . The tangents at T1 and T2 are the lines a2 b2 and x2x + y2y =1 =( 0) a2 b2 . The distances from F c, to these tangents are

1 − cx1 a2 d1(F )= and 2 2 x1 + y1 a4 b4 1 − cx2 a2 d2(F )= . 2 2 x2 + y2 a4 b4 It follows that 2 2 2 x2 + y2 sin T1PF d1(F ) a − cx1 4 4 = = · a b . sin ( ) 2 − 2 2 FPT2 d2 F a cx2 x1 + y1 a4 b4 Similarly, for F =(−c, 0),wehave 2 2 2 x1 + y1 sin T2PF d2(F ) a + cx2 4 4 = = · a b . sin ( ) 2 + 2 2 F PT1 d1 F a cx1 x2 + y2 a4 b4

 sin T1PF sin T2PF =  (2) It is routine to verify that sin FPT2 sin F PT1 ; equivalently,

2 2 2 2 x1 y1 a − cx1 a + cx1 4 + 4 · = a b 2 2 2 2 . a − cx2 a + cx2 x2 + y2 a4 b4 This follows from 2 2 2 2 4 2 2 2 2 x y x 1 x a − c x (a − cx1)(a + cx1) 1 + 1 = 1 + 1 − 1 = 1 = , a4 b4 a4 b2 a2 a4b2 a4b2

2 2 x2 + y2 and an analogous expression for a4 b4 . (3) For convenience, let ∠T1PT2 =Θ, ∠T1PF = θ and ∠T2PF = ϕ.Wehave sin θ sin ϕ = , sin(Θ − θ) sin(Θ − ϕ) sin(Θ − θ)sinϕ =sin(Θ− ϕ)sinθ, cos(Θ − θ − ϕ) − cos(Θ − θ + ϕ)= cos(Θ− θ − ϕ) − cos(Θ − ϕ + θ), cos(Θ − θ + ϕ)= cos(Θ− ϕ + θ).

Since Θ ± (θ − ϕ) <π, we conclude that Θ − θ + ϕ =Θ− ϕ + θ and θ = ϕ. Remark. Similar calculations show that the theorem is also true for hyperbolas and parabola. For a parabola, the line joining P to the “second” focus is parallel to the axis.

Corollary 12.5. The foci of an inscribed conic of a triangle are isogonal conjugate points. 316 Conics Chapter 13

General conics

13.1 Classification of conics

Given five generic points on the Cartesian plane, there is a unique second degree curve C : ax2 +2hxy + by2 +2gx +2fy + c =0 containing the points. We call this the conic determined by the 5 points. A straight line in the plane generically intersects the conic at two points, real or imaginary. When the two points coincide, the line is tangent to the conic. 2 Let Δ:=h − ab be the discriminant of the conic C. Δ =0 bg−hf −hg+af (1) We first assume . By a translation of the origin to the point Δ , Δ , we may assume the conic given by

2 +2 + 2 − J =0 ax hxy by Δ , where ahg 2 2 2 J := abc +2fgh− af − bg − ch = hbf . gfc If J =0, the conic is a pair of straight lines, possibly imaginary. bg−hf −hg+af The point Δ , Δ is the center of the conic. It is clearly the midpoint of the chord cut out by a line through it. 2 2 We begin with a diagonalization of the quadratic form ax +2hxy+by . The symmetric ah matrix always has two real eigenvalues λ1 ≤ λ2 with orthogonal unit eigenvectors. hb The eigenvalues are the roots of the characteristic equation a − λh =0. hb− λ The unit eigenvectors (u, v) and (−v, u) are such that a − λ1 h u a − λ2 h −v =0 and =0. hb− λ1 v hb− λ2 u 318 General conics

These two equations can be combined into a single one: ah u −v u −v λ1 0 = , hb vu vu 0 λ2 and 2 2 ah x ax +2hxy + by = xy hb y u −v λ1 0 uv x = xy vu 0 λ2 −vu y 2 2 = λ1(ux + vy) + λ2(−vx + uy) . 2 Note that λ1λ2 = ab − h = −Δ. The equation of the conic becomes 2 2 λ1Δ(ux + vy) λ2Δ(−vx + uy) + =1. J J = −Δ Δ 0 Δ 0 λ1Δ Since λ1λ2 , this is a hyperbola if > , and an ellipse if < and J and λ2Δ (λ1+λ2)Δ = (a+b)Δ 0 ( + ) 0 J are both positive. This latter condition means J J > , i.e., a b J< . (2) If Δ=0, the above diagonalization still works, though one of the eigenvalues is zero. We write the equation of the conic in the form λ(ux + vy)2 +2gx +2fy + c =0, where λ =0and (u, v) is a unit vector. This can be further rewritten as λ(ux + vy)2 +2g(ux + vy)+2f (−vx + uy)+c =0. If f =0, the conic is a pair of parallel lines (possibly imaginary or coincident). If f =0, we further rewrite the conic equation in the form 2 g λ ux + vy + +2f (−vx + uy + s)=0, λ and recognize it as a parabola.

13.2 Pole and polar We shall identify a point P with coordinates (x, y) with the 1 × 3 matrix xy1 . The conic C can be represented by a symmetric 3 × 3 matrix ⎛ ⎞ ahg M = ⎝hbf⎠ gfc so that a point P lies on the conic if and only if PMPt =0. Two points P and Q are said to be conjugate with respect to C if PMQt =0. Clearly, given C and P , the points conjugate to P with respect to C form a line. This is the polar of P with respect to C. 13.3 Condition of tangency 319

Proposition 13.1. Let P and Q be conjugate with respect to C. If the line PQ intersects the conic at two points, these points divide P and Q harmonically.

Proof. Given P and Q we seek a point λP + μQ with λ + μ =1lying on the conic, i.e.,

0=(λP + μQ)M(λP + μQ)t = λ2PMPt + μ2QMQt since PMQt = QMP t =0. If follows that if X = λP + μQ is one such point, then so is = λP −μQ Y λ−μ . These two points divide P and Q harmonically. Corollary 13.2. If the polar of P intersects C at X and Y , then PX and PY are tangent to C.

13.3 Condition of tangency

We find the condition that a line px + qy + r =0is tangent to the conic C. Assuming q =0. Elimination of y from the equations of the line and the conic leads to the quadratic equation in x:

(aq2 − 2hpq + bp2)x2 +2(gq2 − hqr + brp − fpq)x +(br2 − 2fqr + cq2)=0.

This has a double root if and only if

(gq2 − hqr + brp − fpq)2 − (aq2 − 2hpq + bp2)(br2 − 2fqr + cq2)=0.

Equivalently,

(bc − f 2)p2 +(ac − g2)q2 +(ab − h2)r2 +2(fg − ch)pq +2(hf − bg)pr +2(gh − af)qr =0.

In matrix notation, we rewrite this condition as ⎛ ⎞ ⎛ ⎞ 2 bc − f fg − ch hf − bg p pqr⎝fg − ch ac − g2 gh − af⎠ ⎝q⎠ =0. hf − bg gh − af ab − h2 r

We denote⎛ ⎞ by adj(M) the 3 × 3 matrix in the above equation and call it the adjoint of p M.IfL = ⎝q⎠, the above tangency condition can be expressed as Ltadj(M)L =0. r With respect to the conic, (i) the polar of a point P is the line L = PM, (ii) the pole of a line L is the point P =adj(M)L (normalized so that the third component is 1). If P lies on the conic, its polar is tangent to the conic. If a line L is tangent to the conic, its point of tangency is P =adj(M)L. 320 General conics

13.4 Parametrized conics

Proposition 13.3. A parametrized curve of the form

2 2 a1t + b1t + c1 a2t + b2t + c2 x = ,y= at2 + bt + c at2 + bt + c is a conic.

Proof. Solving for t and t2,wehave

(ac2 − a2c)x +(a1c − ac1)y +(a2c1 − a1c2) t = , (a2b − ab2)x +(ab1 − a1b)y +(a1b2 − a2b1) (b2c − bc2)x +(bc1 − b1c)y +(b1c2 − b2c1) t2 = . (a2b − ab2)x +(ab1 − a1b)y +(a1b2 − a2b1)

It follows that

2 ((ac2 − a2c)x +(a1c − ac1)y +(a2c1 − a1c2))

=((b2c − bc2)x +(bc1 − b1c)y +(b1c2 − b2c1))

· ((a2b − ab2)x +(ab1 − a1b)y +(a1b2 − a2b1)).

Proposition 13.4. A family of lines

p(t)x + q(t)y + r(t)=0 in which p(t), q(t), r(t) are quadratic functions of t envelopes a conic.

Proof. The line intersects its “immediate neighbor” at the point

q(t)r(t) − r(t)q(t) r(t)p(t) − p(t)r(t) x = ,y= . p(t)q(t) − q(t)p(t) p(t)q(t) − q(t)p(t)

Here, both numerators and the common denominator are quadratic functions of t. The result follows from the preceding proposition. 13.5 Rectangular hyperbolas and the nine-point circle 321

13.5 Rectangular hyperbolas and the nine-point circle

Theorem 13.5. (a) A conic through the vertices and the orthocenter of a triangle is a rect- angular hyperbola. (b) The locus of the centers of the rectangular hyperbolas through an orthocentric quadru- ple is the nine-point circle of the quadruple.

A(0,a)

 N  0 − bc H , a

B(b, 0) O C(c, 0)

Proof. (a) Let A =(0,a), B =(b, 0), C =(c, 0). The orthocenter is the point H = 0 − bc , a . Consider a conic through these points, with equation

Ax2 +2hxy + By2 +2gx +2fy + C =0.

Since it contains B and C,

b2A +2bg + C =0, c2A +2cg + C =0.

From these, C (b + c)C A = ,g= − . bc 2bc Since it contains A and H,

a2B − 2af + C =0, b2c2B − 2abcf + a2C =0.

From these, C (a2 − bc)C B = − ,f= . bc 2abc 322 General conics

The equation of the conic is C C (b + c)C (a2 − bc)C · x2 +2hxy − · y2 − · x + · y + C =0. bc bc bc abc a Cx2 − 2bchxy + Cy2 + a(b + c)Cx − (a2 − bc)Cy − abcC =0. To determine the type of the conic, we diagonalize the quadratic form Cx2 − 2bchxy − Cy2. Since the eigenvalues of the symmetric matrix C −bch −bch −C √ are ± C2 − b2c2h2, this conic is a rectangular hyperbola. (b) Indeed, it is possible to find the center of the rectangular hyperbola. By putting x = x + p and y = y + q, we transform the equation of the conic into

a(Cx2 − 2bchxy − Cy2) −a(2Cp +2bchq − (b + c)C)x +(−2abchp +2aCq − (a2 − bc)C)y +a Cp2 − 2bchpq − Cq2 + a(b + c)Cp − (a2 − bc)Cq − abcC =0. We choose p and q such that 2Cp +2bchq − (b + c)C =0, −2abchp +2aCq − (a2 − bc)C =0. The point (p, q) is the center of the rectangular hyperbola. Note that it is possible to eliminate C and h:

(2p − (b + c))C +2bch · q =0, (2aq − (a2 − bc))C − 2bch · ap =0. 2p − (b + c) q =0. 2aq − (a2 − bc) −ap

−2a(p2 + q2)+a(b + c)p +(a2 − bc)q =0. This means that the center (p, q) of the hyperbola lies on the circle + 2 − 2 + 2 − b c · + a bc · =0 x y 2 x 2 y . a b+c − a2−bc This is a circle with center 4 , 4a . It clearly passes through the points (0,0) and the midpoints of BC. = a − b − c =0 If we put y 2 , this becomes x 2 x 2 . Therefore, it passes through the b a c a midpoints 2 , 2 and 2 , 2 of AB and AC. This is the “nine-point circle” of the triangle. 13.5 Rectangular hyperbolas and the nine-point circle 323

Exercise 1. Each of the following conics is a pair of straight lines. Identify the lines and their intersection.

(a) x2 − 2xy − 3y2 − 2x − 6y − 3=0, (b) x2 − y2 − 4x +2y +3=0, (c) 3x2 +6xy +3y2 +2x +2y − 1=0.

2. Find the real intersections of the following pairs of conics.

(a) x2 +2xy − y2 +1=0and x2 +2xy + y2 − 2y =0, (b) x2 +2xy − y2 +1=0and x2 + y2 − 2x − 2y − 1=0, (c) x2 − 2xy − y2 − 1=0and x2 +2xy − y2 − 2y − 1=0, (d) x2 − y2 +1=0and x2 +2xy + y2 − 2x − 1=0, (e) x2 − y2 − 1=0and x2 +2xy + y2 +2x − 2y +1=0, (f) x2 +2xy − y2 +1=0and x2 + y2 − 2x − 1=0, (g) x2 − 2xy − y2 − 1=0and x2 − 2x − 2y =0.

3. Show that the angle ϕ between the pair of straight lines ax2 +2hxy + by2 =0is given by √ Δ tan ϕ = , a + b and that the pair of angle bisectors is given by

hx2 − (a − b)xy − hy2 =0.

x + y =1 4. Find the envelope of the line t a−t . 5. A variable line intersects the x- and y-axes at two points X and Y such that the segment XY has unit length. Is the envelope of the line a conic? x + y =1 6. Find the locus of the pole of the line a b with respect to the variable circles tangent to the coordinate axes and with center on the line x − y =0. 324 General conics Chapter 14

Conic solution of the (O, H, I) problem

14.1 Ruler and compass construction of circumcircle and incircle

Let G and N be the points which divide OH in the ratio

OG : GN : NH =2:1:3.

These are the centroid and the nine-point center of the required triangle (if it exists). A necessary and sufficient condition for the existence of ABC is given by the following theorem.

Theorem 14.1 (Guinand). Let D be the open circular disk with diameter HG. A triangle ABC exists with circumcenter O, orthocenter H, and incenter I if and only if I ∈ D\{N}.

Let R and ρ denote respectively the circumradius and inradius of triangle ABC. From 2 = ( − 2 ) = R − Euler’s formula OI R R r and Feuerbach’s theorem NI 2 r,wehave 2R · NI = OI2. This suggests the following easy ruler-and-compass construction of the circumcircle, nine-point circle, and incircle.

Construction Suppose O, H, I satisfy I ∈ D \{N}. (1) Extend OI to X such that OX =2· OI; construct the circle ONX, and extend NI to intersect the circle again at Y . The length of IY is twice the circumradius, and four times the radius of the nine-point circle. (2) Construct the circumcircle (O) and the nine-point circle (N). (3) Construct the intersection F of the circle (N) with the half line NI, and the circle, center I, passing through F . This is the incircle and F is the (Feuerbach) point of tangency with the nine-point circle.

14.2 The general case: intersections with a conic

Set up a cartesian coordinate system with origin at O. Assume H =(k, 0) and I =(p, q). Since ∠HAI = ∠OAI, and likewise for B and C, the vertices A, B, C are on the locus 326 Conic solution of the (O, H, I) problem

Y

X I F

O N H

Figure 14.1: Incircle from O, H, I of the point P for which ∠HPI = ∠OPI.IfP =(x, y), a routine calculation shows that the locus of P is a curve K: K(x, y)=0, where

K(x, y):=2qx3 − (2p − k)x2y +2qxy2 − (2p − k)y3 −2(p + k)qx2 +2(p2 − q2)xy +2(p − k)qy2 +2kpqx − k(p2 − q2)y. Note that K(k, 0) = 0, i.e., K contains the point H. By computing the circumradius R, we easily obtain the equation of the circumcircle (O): G(x, y)=0, where (p2 + q2)2 G(x, y):=x2 + y2 − . (2p − k)2 +4q2 It is possible to find a linear function L(x, y) such that Q(x, y):=K(x, y) − L(x, y)G(x, y) does not contain third degree terms. For example, by choosing L(x, y):=2qx − (2p − k)y − 2qk, we have

Q(x, y)=−2pqx2 +2(p2 − q2)xy +2pqy2 k2(k − 4p)(p2 − q2)+k(3p2 − 5q2)(p2 + q2)+2p(p2 + q2)2 − · x (2p − k)2 +4q2 2q(kp((2p − k)2 +4q2)+(p2 + q2)2) + · y (2p − k)2 +4q2 2k(p2 + q2)2q − . (2p − k)2 +4q2 14.2 The general case: intersections with a conic 327

The finite intersections of (O) with K are precisely the same with the conic C defined by Q(x, y)=0. Note that the coefficients of x and y in L are dictated by the elimination of the third degree terms in K−L·G. We have chosen the constant term such that L(k, 0) = 0, so that L(x, y)=0represents the line HP parallel to NI. It follows that Q(k, 0) = 0, and the conic C contains the vertices and the orthocenter of the required triangle ABC.It is necessarily a rectangular hyperbola. This fact also follows from the factorization of the quadratic part of Q, namely, −2(px + qy)(qx − py). This means that C is a rectangular q − p hyperbola whose asymptotes have slopes p and q . These are parallel and perpendicular to the segment OI. To construct the rectangular hyperbola C, we identify its center O. This is the point with coordinates (u, v) for which the quadratic polynomial Q(x − u, y − v) has no first degree terms in x and y. A routine calculation gives

k((2p − k)2 − p2 +5q2)+2p(p2 + q2) (p2 + q2 − kp)q O = , 2((2 − )2 +4 2) (2 − )2 +4 2 p k q p k q k (2p − k)(p2 + q2)+2kq2 (p2 + q2)q − kpq = + , . (14.1) 2 2((2p − k)2 +4q2) (2p − k)2 +4q2

Since C is a rectangular hyperbola, its center O lies on the nine-point circle (N). The following observation leads to a very simple construction of the center.

Proposition 14.2. The Feuerbach point F lies on the asymptote perpendicular to OI.

Proof. The nine-point circle has equation

2 k (p2 + q2)2 x − + y2 − =0. (14.2) 2 4((2p − k)2 +4q2)

This intersects the line NI at two points, the Feuerbach point and its antipode (on the nine-point circle). The Feuerbach point is the point k (2p − k)(p2 + q2) (p2 + q2)q F = + , . 2 2((2p − k)2 +4q2) (2p − k)2 +4q2 − p It is easy to see that the line O F has slope q , and is perpendicular to the line OI. 328 Conic solution of the (O, H, I) problem

Corollary 14.3. The center O of the rectangular hyperbola C is the second intersection of the nine-point circle (N) with the perpendicular from F to OI.

F I

O N H O F 

Figure 14.2: Center O of rectangular hyperbola C

It is well known that if C is a rectangular hyperbola passing through the vertices of a tri- angle ABC, its fourth intersection with the circumcircle is the reflection of the orthocenter of the triangle in the center of the hyperbola. Therefore, one of the intersections of C with the circle (O) is the reflection of H in O. The other three are the vertices of the required triangle.

A

F I

O N H O F  C

B

Figure 14.3: Triangle ABC with given O, H, I

14.3 The case OI = IH

If OI = IH, consider the intersection A of the half line NI with the circumcircle. We complete a triangle ABC with (O) as circumcircle and (I) as incircle. The orthocenter of triangle ABC lies on the reflection of AO in the line AI. This is the line AH. Since angle = 1 · = 1 · = R AHN is a right angle, for the midpoint M of AH,wehaveNM 2 AH 2 AO 2 . This means that M is a point on the nine-point circle. It is the midpoint of the segment joining A to the orthocenter. It follows that the orthocenter must be H. 14.3 The case OI = IH 329

Y

A

X I N C O H

B Figure 14.4: Triangle from O, H, I with OI = IH