Chapter 4 Ellipses and Hyperbolas

Chapter 4

Ellipse with two given foci

Given two points F and F in a plane, 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 1

x2 y2 + =1 with b2 := a2 − c2. (4.1) a2 b2

b P a

c a 2 2 O F − a A O F A a F c F c

1 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; (a2 − c2)x2 + a2y2 = a2(a2 − c2). Writing b2 := a2 − c2,wehaveb2x2 + a2y2 = a2b2. 212 Ellipses and Hyperbolas

The directrices and eccentricity of an ellipse

2 =( ) x2 + y =1 2 If P x, y is a point on the ellipse a2 b2 , its square distance from the focus F is 2 | | = c − a PF a x c .

2 2 − a A O F A a c F 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.

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

2 2 2 2 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 . 4.2 Tangents and normals of an ellipse 213

4.2 Tangents and normals of an ellipse

Construction of tangent at P

x1x y1y =( 1 1) + =1 If P x ,y , 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 1 + 2 = = 0 is the line x x y y a . Both tangents intersect the x-axis at T 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

−a θ −c O c a T

−b

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 214 Ellipses and Hyperbolas

Exercise 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

a F O F

2. Prove the following reﬂection property of the ellipse.

2 2 a A O F A a c F c

4.2.1 Evolute of an ellipse

2 x2 + y =1 Consider the ellipse a2 b2 with parametrization x = a cos θ, y = b sin θ, wherea>b(so that the foci are on the x-axis). The center of curvature at P = P (θ) is the a2−b2 · cos3 − a2−b2 · sin3 point a θ, b θ . It can be constructed as follows. (1) Let the tangent at P intersect the axes at X and Y . (2) Complete the rectangle OXQY . (3) Construct the perpendicular from O to the line PQ, to intersect the normal at K. The point K is the center of curvature at P . 4.2 Tangents and normals of an ellipse 215

Q Y

O X

The evolute is the parametric curve

a2 − b2 a2 − b2 x = · cos3 θ, y = − · sin3 θ. a b Eliminating t,wehave 2 2 4 (ax) 3 +(by) 3 = c 3 . This curve is called an astroid.

Q 216 Ellipses and Hyperbolas

4.3 Hyperbolas

Hyperbola with two given foci 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 .

c O a2 a F F c

Assume FF =2c and the constant difference |PF − PF| =2a for a

x2 y2 − =1 with b2 := c2 − a2. (4.2) a2 b2

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 3 | | = 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 3 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 . 4.4 Tangents and normals of a hyperbola 217

B P

θ −c O Q a b c A

4.4 Tangents and normals 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 .

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 218 Ellipses and Hyperbolas

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.

a F O F

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.

4.4.1 Evolute of a hyperbola

2 x2 − y =1 ( )=( sec tan ) For the hyperbola a2 b2 with parametrization x, y a t, b t , the evolute is the parametric curve a2 + b2 a2 + b2 x = · sec3 t, y = − · tan3 t. a b

O Q 4.5 Rectangular hyperbolas 219

4.5 Rectangular hyperbolas

2 Example 4.1. 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 4 1 2 3 4 = −1 ( )= 4 If t is such that t t t t , then it is easy to see that x, y ct , t4 satisﬁes each of the above equations. Therefore, the orthocenter is the point with parameter t4 on the hyperbola.

2 Example 4.2. 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 . Example 4.3. 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 parameters −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.

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

Theorem 4.1. 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. 220 Ellipses and Hyperbolas

F F

=( ) 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 4.7 Miscellaneous exercises 221

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 4.2. The foci of an inscribed conic of a triangle are isogonal conjugate points.

4.7 Miscellaneous exercises

2 ( ) E : x2 + y =1 1. (a) Find the equation of the normal at the point x0,y0 of the ellipse a2 b2 . x + y =1 E 2 2 + 2 2 = (b) Show that if the line p q is normal to the ellipse , then a p b q (a2 − b2)2. 2. 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 3. (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 4. (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.

5. (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 222 Ellipses and Hyperbolas

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.

6. 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.

√1 7. Given an ellipse of eccentricity > 2 , construct two circles with centers on the major axis, each passing through the center and tangent to the ellipse at two points.

8. Given an ellipse, construct an isosceles triangle with apex at a vertex on the minor axis and incircle concentric with the ellipse. 4.7 Miscellaneous exercises 223

9. In a right triangle of sides a and b, an ellipse of semi-axes h, k (parallel to the sides) is inscribed. Show that 2(a − h)(b − k)=ab.

C B 10. An ellipse is tangent to the four sides of a rectangle at their midpoints. Construct the circle tangent to the ellipse and two adjacent sides of the rectangle.

11. Find the locus of the vertices of a rectangular hyperbola with center O and tangent to the line x = a. Hint: Use polar coordinates.

12. 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.

13. Let P be a point on a hyperbola. The intersections of the tangent at P with the asymptotes, and the intersections of the normal at P with the axes lie on a circle through the center of the hyperbola.

a F O F Chapter 5

General conics

5.1 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 226 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.

5.2 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. 5.3 Condition of tangency 227

Proposition 5.1. Let P and Q be conjugate with respect to C. If the line PQintersects 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 5.2. If the polar of P intersects C at X and Y , then PX and PY are tangent to C.

5.3 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. 228 General conics

5.4 Parametrized conics

Proposition 5.3. 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 5.4. 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. 5.5 Rectangular hyperbolas and the nine-point circle 229

5.5 Rectangular hyperbolas and the nine-point circle

Theorem 5.5. (a) A conic through the vertices and the orthocenter of a triangle is a rectangular hyperbola. (b) The locus of the centers of the rectangular hyperbolas through an orthocentric quadruple 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 230 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. 5.5 Rectangular hyperbolas and the nine-point circle 231

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.