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Chapter 33 - PowerPoint PPT Presentation

Chapter 33. hyperbola. 双曲线. 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, e , and greater than one .

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hyperbola

hyperbola

The locusof a point P which moves such that the ratio of its distances from a fixed point Sand from a fixed straight line ZQ is constant, e ,and greater than one.

S is the focus, ZQ the directrix and the e, eccentricity of the hyperbola.

hyperbola

e:eccentricity

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The foci S, S’ are the points (-ae,0), (ae,0) .

Q

Q’

y

The directrices ZQ, Z’Q’ are the lines x=-a/e, x=a/e .

B

A’

A

x

O

Z’

S

Z

S’

AA’ is called the transverse axis=2a .

B’

BB’ is called the conjugate axis=2b .

asymptotes

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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 .

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(i)

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Coordinates of the foci are

(iii) Equations of directrices are

(iv) Equations of asymptotes are

i.e.

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Find the asymptotes of the hyperbola .

Soln:

The asymptotes are

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Find the equation of hyperbola with focus (1,1); directrix 2x+2y=1; e= .

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From definition of a hyperbola, we have PS=ePM .

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.

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Let P be (x,y). Hence, distance from P to (1,1) is :

Distance from P to 2x+2y-1=0 is :

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1. The curve is symmetrical about both axes.

The curve exists for all values of y.

2.

The curve does not exist if |x|<a.

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At the point (a,0) & (-a,0), the gradients are infinite.

4.

Asymptotes of the hyperbola :

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Many results for the hyperbola are obtained from the corresponding results for the ellipse by merely writing in place of .

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1. corresponding results for the ellipse by merely writing in place of . 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

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3. corresponding results for the ellipse by merely writing in place of . The locus of the midpoints of chords of the hyperbola with gradient m is the diameter :

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e.g. 4 corresponding results for the ellipse by merely writing in place of .

Show that there are two tangents to the hyperbola parallel to the line y=2x-3 and find their distance apart.

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Soln: corresponding results for the ellipse by merely writing in place of .

Hence, equations of tangents are :

Perpendicular distance from (0,0) to the lines are :

O

Distance=

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The rectangular hyperbola corresponding results for the ellipse by merely writing in place of .

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1. corresponding results for the ellipse by merely writing in place of . A hyperbola with perpendicular asymptotes is a rectangular hyperbola.

i.e. b=a

So, the standard equation of a rectangular hyperbola is :

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2. corresponding results for the ellipse by merely writing in place of . Eccentricity of a rectangular hyperbola =

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y asymptotes

y

x

O

x

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y

y

xy=-9

xy=9

x

o

x

o

y

y

x

x

o

o

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hyperbola asymptotes

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The equation, asymptotes

is satisfied if

t is a parameter.

The parametric coordinates of any point are :

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Tangent and normal at the point asymptotes(ct,c/t) to the curve

Gradient of tangent at (ct,c/t) is

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Equation of tangent at asymptotes(ct,c/t) is

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Equation of normal at asymptotes(ct,c/t) is

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e.g. 5 asymptotes

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.

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Soln: asymptotes

Let P be the point (2t,2/t).

Equation of tangent at P is

x-axis and y-axis are the asymptotes.

When y=0,Q is (4t,0), when x=0 R is (0,4/t).

The midpoint of QR is (2t,2/t).

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e.g. 6 asymptotes

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.

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Soln: asymptotes

Let S(ct,c/t), R(cp,c/p) and P(cq,c/q).

Gradient of tangent at P is :

S

R

P

at

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Hence,

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Conclusion: asymptotes

In analytic geometry, the hyperbola is represented by the implicit equation :

The condition : B2 − 4AC > 0

• (if A + C = 0, the equation represents a rectangular hyperbola. )

Ellipse

In the asymptotesCartesian 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 :

Ax2 + Bxy + Cy2 + Dx + Ey + F = 0with A, B, Cnot all zero.

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then: asymptotes

• if B2 − 4AC < 0, the equation represents an ellipse (unless the conic is degenerate, for example x2 + y2 + 10 = 0);

• ifA = C and B = 0, the equation represents a circle;

• if B2 − 4AC = 0, the equation represents a parabola;

• if B2 − 4AC > 0, the equation represents a hyperbola;

• (if A + C = 0, the equation represents a rectangular hyperbola. )

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Analyzing an Hyperbola asymptotes

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 4x2 - 9y2 + 32x + 18y + 91 = 0.

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~ The end ~ asymptotes

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Ex 14d asymptotesdo Q1, 3, 5, 7, 9, 11.

Misc.14 no need to do

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Ex 14d Q 1 asymptotes

At point (2a,a/2),

Gradient of normal at (2a,a/2) is 4.

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Equation of normal at asymptotes(2a,a/2) is :

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Ex 14d Q 3 asymptotes

(2t,2/t)

2y=x+7

4/t=2t+7

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At point asymptotesA, t=1/2 so, A is (1,4)

At point B, t=-4 so, B is (-8,-1/2)

x=2t, y=2/t

Hence, xy=4

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At point asymptotesA, dy/dx=-4

At point B,dy/dx=-1/16

Eqn of tangent at A :

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Eqn of tangent at asymptotesB :

Hence, 16(8-4x)+x+16=0

128-64x+x+16=0

x=144/63=16/7

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x=16/7 asymptotes

Therefore, y=8-4(16/7)

=8-64/7

=-8/7

Point of intersection is

(16/7, -8/7)

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Ex 14d Q 5 asymptotes

Let the point on the curve be (ct,c/t) .

4xy=25

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Eqn of tangent lines at asymptotes(ct,c/t) :

3t -1

t 2

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(3t-1)(t+2)=0 asymptotes

t=1/3 or t=-2

When t=1/3,

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When asymptotest=-2,

So, m1=-1/4 and m2=-9

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Let the two angles of these tangent lines be asymptotesA and B.

m1=-1/4 and m2=-9

Hence,

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Ex 14d Q 5 asymptotes

Why this method can’t be accepted???

(2,-3) 4xy=25

Equation of tangent lines in gradient form :

Therefore,

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Let asymptotesm1 and m2 be the roots,

We have , and

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Let the two angles between these tangent lines be asymptotesA and B.

---------------------1

We need to know m1-m2 now.

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Sub. Into asymptotes1

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Ex 14d Q 7 asymptotes

xy=9 (-1,-9)

Eqn of normal line at (-1,-9) :

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Hence, asymptotes

Another point is y=1/9, x=81

So, length of the chord is

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Ex 14d Q 9 asymptotes

To find points of intersection of 2 hyperbola :

----------- 1

Sub. into 1

and

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At asymptotes(1,4)

The product of these gradient =-1.

At (-1,-4)

Hence, the product of these gradient =-1.

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y asymptotes

Ex 14d Q 11

Q

P(2,9)

y(x-1)=9

x

R

2

At P, y’=-9

Eqn of tangent line at P, (y-9)=(-9)(x-2)

y=-9x+27

At Q, x=1, y=18

At R, y=0, x=3

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P(2,9), Q(1,18), R(3,0) asymptotes

Hence, QP=PR

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The 2009 year end examination scopes : asymptotes

• The straight line

• The circle

• The parabola

• The ellipse

• The hyperbola 排列与组合

• Permutations & combinations

• Probability概率

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2 asymptotesnd November, 2009 (Monday) S2S Mathematics Final Examination

Part A : Short Questions -- answer all 9 questions x 5%=45%

Part B : Long Questions -- answer 5 from 9questions x 11%=55%

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