Chapter 14 hyperbola 3102021 hyperbola 1 3102021 hyperbola
Chapter 14 hyperbola 双曲� 3/10/2021 hyperbola 1
3/10/2021 hyperbola 2
Definition: The locus of a point P which moves such that the ratio of its distances from a fixed point S and from a fixed straight line ZQ is constant and greater than one. S is the focus, ZQ the directrix and the e, eccentricity of the hyperbola. 3/10/2021 hyperbola 3
Simplest form of the eqn of a hyperbola e: eccentricity 3/10/2021 hyperbola 4
Q The foci S, S’ are the points (-ae, 0), (ae, 0). Q’ y The directrices ZQ, Z’Q’ are the lines x= A’ A x -a/e, x=a/e. O Z’ S’ S Z AA’ is called the 实轴 transverse axis=2 a. B’ B BB’ is called the 虚轴 conjugate axis=2 b. asymptotes 3/10/2021 hyperbola 5
e. g. 1 For the hyperbola , find (i) the eccentricity, (ii) the coordinates of the foci (iii) the equations of the directrices and (iv) the equations of the asymptotes. 3/10/2021 hyperbola 6
Soln: (i) hyperbola 7 3/10/2021
(ii) Coordinates of the foci are (iii) Equations of directrices are (iv) Equations of asymptotes are i. e. 3/10/2021 hyperbola 8
e. g. 2 Find the asymptotes of the hyperbola . Soln: The asymptotes are 3/10/2021 hyperbola 9
e. g. 3 Find the equation of hyperbola with focus (1, 1); directrix 2 x+2 y=1; e=. 3/10/2021 hyperbola 10
Soln: From definition of a hyperbola, we have PS=e. PM. Where PS is the distance from focus to a point P and PM is the distance from the directrix to a point P. e is the eccentricity of the hyperbola. 3/10/2021 hyperbola 11
Let P be (x, y). Hence, distance from P to (1, 1) is : Distance from P to 2 x+2 y-1=0 is : 3/10/2021 hyperbola 12
is the required equation of the hyperbola. 3/10/2021 hyperbola 13
14 Properties of the hyperbola 3/10/2021
1. The curve is symmetrical about both axes. The curve exists 2. for all values of y. The curve does not exist if |x|<a. 3/10/2021 hyperbola 15
3. At the point (a, 0) & (-a, 0), the gradients are infinite. 4. Asymptotes of the hyperbola : 3/10/2021 hyperbola 16
Many results for the hyperbola are obtained from the corresponding results for the ellipse by merely writing in place of. 3/10/2021 hyperbola 17
1. The equation of the tangent to the hyperbola at the point (x’, y’) is 2. The gradient form of the equation of the tangent to the hyperbola is 3/10/2021 hyperbola 18
3. The locus of the midpoints of chords of the hyperbola with gradient m is the diameter : 3/10/2021 hyperbola 19
e. g. 4 Show that there are two tangents to the hyperbola parallel to the line y=2 x-3 and find their distance apart. 3/10/2021 hyperbola 20
Soln: Gradient of tangents=2 Hence, equations of tangents are : O 3/10/2021 Perpendicular distance from (0, 0) to the lines are : Distance= hyperbola 21
The rectangular hyperbola 22 hyperbola 3/10/2021
1. A hyperbola with perpendicular asymptotes is a rectangular hyperbola. i. e. b=a So, the standard equation of a rectangular hyperbola is : 3/10/2021 hyperbola 23
2. Eccentricity of a rectangular hyperbola = 3/10/2021 hyperbola 24
Equation of a rectangular hyperbola with respect to its asymptotes 3/10/2021 hyperbola 25
y y x O x 3/10/2021 hyperbola 26
Some simple sketches of the rectangular hyperbola : y xy=-9 xy=9 x o y y y o 3/10/2021 x o x x o hyperbola 27
3/10/2021 hyperbola 28
Parametric equations of a rectangular hyperbola 3/10/2021 hyperbola 29
The equation, is satisfied if t is a parameter. The parametric coordinates of any point are : 3/10/2021 hyperbola 30
Tangent and normal at the point (ct, c/t) to the curve Gradient of tangent at (ct, c/t) is 3/10/2021 hyperbola 31
Equation of tangent at (ct, c/t) is 3/10/2021 hyperbola 32
Equation of normal at (ct, c/t) is 3/10/2021 hyperbola 33
e. g. 5 The tangent at any point P on the curve xy=4 meets the asymptotes at Q and R. Show that P is the midpoint of QR. 3/10/2021 hyperbola 34
Soln: Let P be the point (2 t, 2/t). Equation of tangent at P is x-axis and y-axis are the asymptotes. When y=0, Q is (4 t, 0), when x=0 R is (0, 4/t). The midpoint of QR is (2 t, 2/t). 3/10/2021 hyperbola 35
Miscellaneous examples on the hyperbola 3/10/2021 hyperbola 36
e. g. 6 A chord RS of the rectangular hyperbola subtends a right angle at a point P on the curve. Prove that RS is parallel to the normal at P. 3/10/2021 hyperbola 37
Soln: Let S(ct, c/t), R(cp, c/p) and P(cq, c/q). S Gradient of tangent at P is : R P at 3/10/2021 hyperbola 38
Gradient of PS= Gradient of RP= Gradient of RS= Hence, 3/10/2021 hyperbola 39
Conclusion: In analytic geometry, the hyperbola is represented by the implicit equation : The condition : B 2 − 4 AC > 0 (if A + C = 0, the equation represents a rectangular hyperbola. ) 3/10/2021 Ellipse 40
In the Cartesian coordinate system, the graph of a quadratic equation in two variables is always a conic section, and all conic sections arise in this way. The equation will be of the form : 2 Ax + Bxy + + Dx + Ey + F = 0 with A, B, C not all zero. 3/10/2021 2 Cy hyperbola 41
then: • if B 2 − 4 AC < 0, the equation represents an ellipse (unless the conic is degenerate, for example x 2 + y 2 + 10 = 0); • if A = C and B = 0, the equation represents a circle; • if B 2 − 4 AC = 0, the equation represents a parabola; • if B 2 − 4 AC > 0, the equation represents a hyperbola; (if A + C = 0, the equation represents a rectangular hyperbola. ) 3/10/2021 hyperbola 42
Analyzing an Hyperbola State the coordinates of the vertices, the coordinates of the foci, the lengths of the transverse and conjugate axes and the equations of the asymptotes of the hyperbola defined by 4 x 2 - 9 y 2 + 32 x + 18 y + 91 = 0. 3/10/2021 hyperbola 43
~ The end ~ 3/10/2021 hyperbola 44
Ex 14 d do Q 1, 3, 5, 7, 9, 11. Misc. 14 no need to do 45 hyperbola 3/10/2021
Ex 14 d Q 1 At point (2 a, a/2), Gradient of normal at (2 a, a/2) is 4. 3/10/2021 hyperbola 46
Equation of normal at (2 a, a/2) is : 3/10/2021 hyperbola 47
Ex 14 d Q 3 (2 t, 2/t) 2 y=x+7 4/t=2 t+7 3/10/2021 hyperbola 48
At point A, t=1/2 so, A is (1, 4) At point B, t=-4 so, B is (-8, -1/2) x=2 t, y=2/t Hence, 3/10/2021 xy=4 hyperbola 49
At point A, dy/dx=-4 At point B, dy/dx=-1/16 Eqn of tangent at A : 3/10/2021 hyperbola 50
Eqn of tangent at B : Hence, 3/10/2021 16(8 -4 x)+x+16=0 128 -64 x+x+16=0 x=144/63=16/7 hyperbola 51
x=16/7 Therefore, y=8 -4(16/7) =8 -64/7 =-8/7 Point of intersection is (16/7, -8/7) 3/10/2021 hyperbola 52
Ex 14 d Q 5 Let the point on the curve be (ct, c/t). 4 xy=25 3/10/2021 hyperbola 53
Eqn of tangent lines at (ct, c/t) : 3/10/2021 hyperbola 3 t -1 t 2 54
(3 t-1)(t+2)=0 t=1/3 or t=-2 When t=1/3, 3/10/2021 hyperbola 55
When t=-2, So, m 1=-1/4 and m 2=-9 3/10/2021 hyperbola 56
Let the two angles of these tangent lines be A and B. m 1=-1/4 and m 2=-9 Hence, 3/10/2021 hyperbola 57
Ex 14 d Q 5 (2, -3) Why this method can’t be accepted? ? ? 4 xy=25 Equation of tangent lines in gradient form : Therefore, 3/10/2021 hyperbola 58
Let m 1 and m 2 be the roots, We have , 3/10/2021 and hyperbola 59
Let the two angles between these tangent lines be A and B. -----------1 We need to know m 1 -m 2 now. 3/10/2021 hyperbola 60
Sub. Into 1 3/10/2021 hyperbola 61
Hence, the acute angle is 3/10/2021 hyperbola . 62
Ex 14 d Q 7 xy=9 (-1, -9) Eqn of normal line at (-1, -9) : 3/10/2021 hyperbola 63
Hence, Another point is y=1/9, x=81 So, length of the chord is 3/10/2021 hyperbola 64
Ex 14 d Q 9 To find points of intersection of 2 hyperbola : ------ 1 Sub. into 1 and 3/10/2021 hyperbola 65
At (1, 4) The product of these gradient =-1. At (-1, -4) Hence, the product of these gradient =-1. 3/10/2021 hyperbola 66
y Ex 14 d Q 11 Q y(x-1)=9 2 At P, y’=-9 Eqn of tangent line at P, At Q, x=1, y=18 At R, y=0, x=3 3/10/2021 P(2, 9) hyperbola R x (y-9)=(-9)(x-2) y=-9 x+27 67
P(2, 9), Q(1, 18), Hence, 3/10/2021 R(3, 0) QP=PR hyperbola 68
The 2009 year end examination scopes : 1. The straight line 2. The circle 3. The parabola 4. The ellipse 5. The hyperbola 排列与组合 6. Permutations & combinations 7. Probability概率 3/10/2021 hyperbola 69
nd 2 November, 2009 (Monday) S 2 S Mathematics Final Examination Part A : Short Questions -- answer all 9 questions x 5%=45% Part B : Long Questions -- answer 5 from 9 questions x 11%=55% 3/10/2021 hyperbola 70
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