1 definition of a hyperbola hyperbolas problem 4 problem 1 standard 4, 9, 16, 17 problem 3 problem 2...

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DEFINITION OF A HYPERBOLA

HYPERBOLAS

PROBLEM 4

PROBLEM 1

Standard 4, 9, 16, 17

PROBLEM 3

PROBLEM 2

STANDARD FORMULAS FOR HYPERBOLAS

PROBLEM 5

END SHOW

ALL CONICS IN ONE EQUATION

PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

http://www.mrperezonlinemathtutor.com/

2

STANDARD 4:

Students factor polynomials representing the difference of squares, perfect square trinomials, and the sum and difference of two cubes

STANDARD 9:

Students demonstrate and explain the effect that changing a coefficient has on the graph of quadratic functions; that is, students can determine how the graph of a parabola changes as a, b, and c vary in the equation y = a(x-b) + c.

STANDARD 16:

Students demonstrate and explain how the geometry of the graph of a conic section (e.g., asymptotes, foci, eccentricity) depends on the coefficients of the quadratic equation representing it.

STANDARD 17:

Given a quadratic equation of the form ax + by + cx + dy + e = 0, students can use the method for completing the square to put the equation into standard form and can recognize whether the graph of the equation is a circle, ellipse, parabola, or hyperbola. Students can then graph the equation.

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2 2

ALGEBRA II STANDARDS THIS LESSON AIMS:



ESTÁNDAR 4:

Los estudiantes factorizan polinomios representando diferencia de cuadrados, trinomios cuadrados perfectos, y la suma de diferencia de cubos.

ESTÁNDAR 9:

Los estudiantes demuestran y explican los efectos que tiene el cambiar coeficientes en la gráfica de funciones cuadráticas; esto es, los estudiantes determinan como la gráfica de una parábola cambia con a, b, y c variando en la ecuación y=a(x-b) + c

ESTÁNDAR 16:

Los estudiantes demuestran y explican cómo la geometría de la gráfica de una sección cónica (ej. Las asimptótes, focos y excentricidad) dependen de los coeficientes de la ecuación cuadrática que las representa.

Estándar 17:

Dada una ecuación cuadrática de la forma ax +by + cx + dy + e=0, los estudiantes pueden usar el método de completar al cuadrado para poner la ecuación en forma estándar y pueden reconocer si la gráfica es un círculo, elipse, parábola o hipérbola. Los estudiantes pueden graficar la ecuación

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22



4

Standard 4, 9, 16, 17HYPERBOLA

Definition of Hyperbola:

A hyperbola is the set of all points in a plane such that the absolute value of the difference of the distances from any point on the hyperbola to two given point, called foci, is constant.

d1 d2

d3 d4

d5 d6-

-

-

= k

= k

= k

x

y

F1

F2

d2d1

d4d3

d5 d6



5

Standard 4, 9,16, 17

PARTS OF A HYPERBOLA

x

y

F1

F2

asymptote asymptote

vertex vertex

b

a

ctransverse axis

con

juga

te a

xis



6

Standard 4, 9, 16, 17STANDARD EQUATIONS OF A HYPERBOLA

• Hyperbola with center at (h,k) with horizontal axis

has equation

In this case, transverse axis is horizontal.

• Hyperbola with center at (h,k) with vertical axis

has equation

In this case, transverse axis is vertical.

NOTE: These two hyperbolas are graphed with center (0,0)

(x – h) (y – k)2 2= 1

a2 b2-

(y – k) (x – h)2 2= 1

b2a2 -

c = a + b2 2 2 For both equations.

x

y

F1

F2

y

x

F1

F2

7

Standard 4, 9, 16, 17

Given the graph below obtain the equation of the hyperbola.

42 6-2-4-6

2

4

6

-2

-4

-6

8 10-8-10

8

-8

10

x

y

13(2+ , 0)13(2- , 0)(-1,0)

(5,0)

Transverse axis is 6 units:

2a=62 2

a = 3 a = 92

c = a + b2 2 2

-a -a2 2

b = c - a2 22

Focus1 =( h+c,k)

= 13(2+ , 0)

Center = (2,0)

From the figure:

Focus2 Focus1

then

If

2 + c = 2 + 13

2+c

-2 -2c = 13

2+c

b =22 13 -9

b = 13-92

b = 42

(x – h) (y – k)2 2= 1

a2 b2-

h= 2k= 0

(x-(+2)) (y-(0))2 2

=1 9 4

-

(x-2) y2 2

=1 9 4

-

(2,0)

Hyperbola is horizontal:

then we know:



8

Standard 4, 9,16, 17Given the graph below obtain the equation of the hyperbola.

42 6-2-4-6

2

4

6

-2

-4

-6

8 10-8-10

8

-8

10

x

y

13(4,-2- )

13(4,-2+ )

Focus2

Focus1

(4,-2)

Focus1 =( h,k+c)

= 13(4,-2+ )

Center = (4,-2)

From the figure:

then

If

-2 + c = -2 + 13

-2+c

+2 +2c = 13

-2+c

h= 4k= -2

Transverse axis is 4 units:

2a=42 2

a = 2 a = 42

c = a + b2 2 2

-a -a2 2

b = c - a2 22

b =22 13 -4

b = 13-42

b = 92

then we know:

(y – k) (x – h)2 2= 1

a2 b2- (y-(-2)) (x-(4))2 2

=14 9

-

Hyperbola is vertical:

(y+2) (x-4)2 2

=1 4 9

-



9

Standard 4, 9, 16, 17Draw the hyperbola that is represented by:

(x-3) (y-4)2 2

=125 9

-

(x -(+3)) (y-(+4))2 2

=125 9

-

a = 5a = 252

b = 92 b= 3

c = a + b2 2 2

c = 25 + 92

c = 342

c= 34 5.8

Focus 1= (h+c, k)

Focus 2= (h-c, k)

= (3+5.8,4)= (8.8,4)= (3-5.8,4)= (-2.8,4)

Center = (3,4)h= 3k= 4

42 6-2-4-6

2

4

6

-2

-4

-6

8 10-8-10

8

-8

10

x

y



10

Standard 4, 9, 16, 17Draw the hyperbola that is represented by:

(y-6) (x+2)2 2

=1 81 64

-

(y-(+6)) (x-(-2))2 2

=181 64

-

a = 9a = 812

b = 642 b= 8

c = a + b2 2 2

c = 81 + 642

c = 1452

c= 145 12

Focus 1= (h, k+c)

Focus 2= (h, k–c )

= (-2,6+12)= (-2,18)= (-2,6-12)= (-2,-6)

Center = (-2,6)h= -2k= 6

y

84 12-4-8-12

4

8

12

-4

-8

-12

16 20-16-20

16

-16

20

x



11

Standard 4, 9,16, 1736x - 25y -144x +150y - 981 = 0 2 2

36x -144x + -25y +150y -981 = 02 2

36x -36(4)x + -25y -(-25)6y - 981 = 02 2

36 x - 4x + - 25 y - 6y + - 981 = 36 - 252 242

2 42

262

2 62

2

36 x - 4x + - 25 y - y + - 981 = 36 - 252 2 (2)2

(2)2 (3)

2(3)

2

4 (4)9 (9)36 x - 4x + - 25 y - 6y + - 981 = 36 - 252 2

36(x-2) - 25(y-3) - 981 = 144 - 2252 2

36(x-2) - 25 (y-3) - 981 = -812 2

+981 +981

36(x-2) - 25(y-3) = 9002 2

900 900

We know that is an hyperbola. Put it in the standard form , graph it, and finally find the equation

of the asymptotes.

(x – h) (y – k)2 2= 1

a2 b2-

36x - 25y -144x +150y - 981 = 0 2 2



12


36(x-2) - 25(y-3) = 9002 2

900 900

36(x-2) 25(y-3)2 2

=1900 900

-

36(x-2) 25(y-3)2 2

=1900 90036

36 25

25

-

(x-2) (y-3)2 2

=125 36

-

(x-(+2)) (y-(+3))2 2

=125 36

-

(x – h) (y – k)2 2= 1

a2 b2-

h= 2

k= 3

a = 5a = 252

b = 362 b= 6

c = a + b2 2 2

c = 25 + 362

c = 612

c= 61 7.8

Focus 1= (h+c, k)

Focus 2= (h-c, k)

= (2+7.8,3)= (9.8,3)= (2-7.8,3)= (-5.8,3)

42 6-2-4-6

2

4

6

-2

-4

-6

8 10-8-10

8

-8

10

x

yCenter = (2,3)

36x - 25y -144x +150y - 981 = 0 2 2

We know that is an hyperbola. Put it in the standard form , graph it, and finally find the equation

of the asymptotes.

(x – h) (y – k)2 2= 1

a2 b2-



13


42 6-2-4-6

2

4

6

-2

-4

-6

8 10-8-10

8

-8

10

x

y

Equations for the asymptotes.

Center = (2, 3)

-5

m=5-

6m=

5

6+

5+

6+

6+=

65

-

(y – ) = m(x – )x1

y1

(y – ) = (x – )

x1

2

y1

365

(y – ) = (x – )23 65

-

y – 3 = (x – 2)65

-y - 3 = (x – 2)65

y - 3 = x - 12 5

65

y – 3 = x +12 5

65

-

3 55

= 15 5

y - = x - 12 5

65

15 5

y - = x + 12 5

65

-15 5

+ 15 5

+ 15 5 + 15

5+ 15

5

y = x + 3 5

65

y = x + 27 5

65

-

First we find the slope for both asymptotes:

Using the center and the point-slope form of the equation of a line:


14

Standard 4, 9, 16, 17

Equation of a Conic Section

Ax + Bxy + Cy + Dx + Ex + F = 02 2

Conic Section Relationship of A and C

Parabola A=0 or C=0, but not both

Circle A=C

EllipseA and C have the same sign and A=C

Hyperbola A and C have opposite signs

Now let’s use the exercises we solved before!


15

Standard 4, 9, 16, 17Match the following equations with the corresponding graph:

Ax + Bxy + Cy + Dx + Ex + F = 02 2

16x + 64y -64x -384y + 384 = 02 2

x

y

4x + 24x – y +16 =02

y

x + y + 8x +2y -32=0 2 2

x

y

36x - 25y -144x +150y - 981 = 0 2 2

x

yC=1

C=+64

C=0

C= -25

A=1

A=+16

A=4

A=+36

4)

1)

2)

3)

1

2

3

4


16

42 6-2-4-6

2

4

6

-2

-4

-6

8 10-8-10

8

-8

10

x

y

Standard 4, 9, 16, 17Solve the following system of equations:

y = x – 2x – 3 2

y = -3x – 1

Graphing the parabola:

y = x – 2x – 3 2

y = x – 2x + 1 – 3 – 1 2

y = (x – 1) – 4 2

h= 1

k= - 4

Vertex: (1, - 4)

Axis of symmetry: x= 1

a= 1

Latus rectum: 1

Focus: ( 1, - 4 + )1

4 1

= ( 1, 3.75)

Directrix: y = -4 - 1

4( )1= - 4.25

Graphing the line:

y = -3x – 1

m = -3 =-3+1

b= -1

Now let’s check this result algebraically!

x= 1

y = -4.25

(-2, 5)

(1,- 4)

The solution is:

(-2, 5) and (1, - 4)PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

17

42 6-2-4-6

2

4

6

-2

-4

-6

8 10-8-10

8

-8

10

x

y

Standard 4, 9, 16, 17Solve the following system of equations:

y = x – 2x – 3 2

y = -3x – 1 x= 1

y = -4.25

(-2, 5)

(1,- 4)

Solving by substitution:

= x – 2x – 3 2 -3x – 1

+1 +1

-3x = x – 2x – 2 2

+3x +3x

0 = x – x – 2 2

0 = (x + 2)(x – 1)

-2 1

(2)(-1) 2 + -1= 1

x + 2= 0 x – 1= 0

-2 -2x = -2

+1 +1

x = 1

y = -3( ) – 1 -2

y = 6 – 1

y = 5

Using: x= -2

y = -3( ) – 1 1

y = -3 – 1

y = -4

Using: x= 1

(-2, 5) (1, - 4)PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

1 definition of a hyperbola hyperbolas problem 4 problem 1 standard 4, 9, 16, 17 problem 3 problem 2...

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