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Chapter 14 Hyperbolas

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Chapter 14 Hyperbolas Chapter 14 Hyperbolas 14.1 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 ′. b P c ′ O a2 a F F c Assume FF ′ = 2c and the constant difference PF PF ′ = 2a for a<c. Setupa | − | coordinate system such that F =(c, 0) and F ′ =( c, 0). A point (x,y) is on the hyperbola if and only if − (x + c)2 + y2 (x c)2 + y2 =2a. | − − | p p This can be reorganized as x2 y2 =1 with b2 := c2 a2. (14.1) a2 − b2 − 320 Hyperbolas The directrices and eccentricity of a hyperbola x2 y2 If P = (x,y) is a point on the hyperbola a2 b2 = 1, its square distance from the focus 2 − F is 1 PF = c x a . Writing e := c , we have PF = e x a . The hyperbola can | | a − c a · − e also be regarded as the locus of point P whose distances from F (focus) and the line x = a e 2 a (directrix) bear a constant ratio e> 1 (the eccentricity). Similarly, PF ′ = e x + c . The a | | point F ′ = ( ae, 0) and the line x = form another pair of focus and directrix of the − − e same hyperbola. There is an easy parametrization of the hyperbola: P (θ)=(a sec θ, b tan θ). B P T θ c O Q a b c A − 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 PQ is tangent to the hyperbola. 14.2 Tangents and normals of a hyperbola x2 y2 Let P be a point on the hyperbola H : 2 2 = 1. The circles with diameters PF and a − b 1 PF2 are tangent to the auxiliary circle C (O,a). The line joining the points of tangency is tangent to H at P . 2 2 2 1 2 2 2 2 2 x b 2 2 2 c 2 2 PF =(x c) + y =(x c) + b 1 a2 = 1 a2 x 2cx + c + b = a2 x 2cx + a = 2 − − − − − − c x a . a − 14.2 Tangents and normals of a hyperbola 321 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. P ′ a F O F a2+b2 a2+b2 The center of the circle is 2a sec θ, 2b tan θ . The locus is the hyperbola 4(a2x2 b2y2)=(a2 + b2)2. − 322 Hyperbolas 14.2.1 Evolute of a hyperbola x2 y2 For the hyperbola a2 b2 =1 with parametrization (x,y)=(a sec t, b tan t), the evolute is the parametric curve− a2 + b2 a2 + b2 x = sec3 t, y = tan3 t. a · − b · P O Q 14.3 Rectangular hyperbolas 323 14.3 Rectangular hyperbolas 2 The line joining the points t, t′ of the rectangular hyperbola xy = c is x + tt′y c(t + t′)=0. − 2 Example 14.1. Four points on the rectangular hyperbola xy = c with parameters t1, t2, t , t form an orthocentric system if and only if t t t t = 1. 3 4 1 2 3 4 − 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 t t t x t y + c(1 t2t t )= 0, 1 2 3 − 1 − 1 2 3 t t t x t y + c(1 t t2t )= 0, 1 2 3 − 2 − 1 2 3 t t t x t y + c(1 t t t2)= 0. 1 2 3 − 3 − 1 2 3 If t is such that t t t t = 1, then it is easy to see that (x,y) = ct , c satisfies each 4 1 2 3 4 − 4 t4 of the above equations. Therefore, the orthocenter is the point with parameter t4 on the hyperbola. 2 Example 14.2. 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 c2 2cf c2t2 + +2cgt + + c =0, t2 t 0 or 2 4 3 2 2 c t +2cgt + c0t +2cft + c =0. c2 From this we conclude that t1t2t3t4 = c2 =1. Example 14.3. PQ is 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, t , t , t are concyclic, τ t t t =1, and t t −t τ = 1. This means that P is − 1 2 3 − · 1 2 3 1 2 3 · − the orthocenter of the triangle with vertices t1, t2, t3. This coincides with the circumcenter, and the triangle is equilateral. 324 Hyperbolas Exercise 1. The pedal of the point p,q on the line joining ct , c and ct , c is the point ( ) 1 t1 2 t2 t2t2p t t q +(t + t )c t t p + q + t t (t + t )c 1 2 − 1 2 1 2 , − 1 2 1 2 1 2 . 1+ t2t2 1+ t2t2 1 2 1 2 2. 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 ′. 14.4 A theorem on the tangents from a point to a conic Theorem 14.1. 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. P b T1 b T2 b F ′ F b b x1x y1y Proof. (1) Let T1 = (x1, y1). The tangents at T1 and T2 are the lines a2 + b2 = 1 and x2x y2y a2 + b2 =1. The distances from F =(c, 0) to these tangents are cx1 1 2 d F − a and 1( )= 2 2 x1 y1 a4 + b4 cx2 q1 2 d F − a . 2( )= 2 2 x2 y2 a4 + b4 q 14.5 Miscellaneous exercises 325 It follows that 2 2 x2 y2 sin T PF d (F ) a2 cx 4 + 4 1 = 1 = 1 a b . 2 − 2 2 sin FPT2 d2(F ) a cx2 · q x1 y1 − a4 + b4 q Similarly, for F ′ =( c, 0), we have − 2 2 x1 y1 sin T PF d (F ) a2 + cx 4 + 4 2 ′ = 2 ′ = 2 a b . 2 2 2 sin F ′PT1 d1(F ′) a + cx1 · q x2 y2 a4 + b4 ′ q sin T1P F sin T2P F (2) It is routine to verify that ′ ; equivalently, sin FPT2 = sin F PT1 2 2 2 2 x1 y1 a cx1 a + cx1 a4 + b4 − = 2 2 . a2 cx a2 cx x2 y2 2 · + 2 4 + 4 − a b This follows from x2 y2 x2 1 x2 a4 c2x2 (a2 cx )(a2 + cx ) 1 + 1 = 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. We have sin θ sin ϕ = , sin(Θ θ) sin(Θ ϕ) − − sin(Θ θ)sin ϕ = sin(Θ ϕ)sin θ, − − 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 14.2. The foci of an inscribed conic of a triangle are isogonal conjugate points. 14.5 Miscellaneous exercises E x2 y2 1. (a) Find the equation of the normal at the point (x0, y0) of the ellipse : a2 + b2 =1. x y E 2 2 2 2 (b) Show that if the line p + q = 1 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.
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