HYPERBOLAS Standard 4, 9, 16, 17 DEFINITION OF A HYPERBOLA

Presentación del tema: "HYPERBOLAS Standard 4, 9, 16, 17 DEFINITION OF A HYPERBOLA"— Transcripción de la presentación:

1 HYPERBOLAS Standard 4, 9, 16, 17 DEFINITION OF A HYPERBOLA
STANDARD FORMULAS FOR HYPERBOLAS PROBLEM 1 PROBLEM 2 PROBLEM 3 PROBLEM 4 PROBLEM 5 ALL CONICS IN ONE EQUATION END SHOW PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

2 ALGEBRA II STANDARDS THIS LESSON AIMS:
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. 2 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

3 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 2 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

4 Standard 4, 9, 16, 17 x d d d d d - d = k d - d = k d d d - d = k
HYPERBOLA 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. x y F 1 2 d 3 d 4 d 1 d 2 d 1 - d 2 = k d 3 - d 4 = k d 5 d 6 d 5 - d 6 = k PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

5 Standard 4, 9,16, 17 PARTS OF A HYPERBOLA y asymptote asymptote vertex
2 asymptote asymptote vertex vertex conjugate axis b c transverse axis a PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

6 STANDARD EQUATIONS OF A HYPERBOLA
x y F 1 2 Hyperbola with center at (h,k) with horizontal axis has equation (x – h) (y – k) 2 = 1 a b - In this case, transverse axis is horizontal. y x F 1 2 Hyperbola with center at (h,k) with vertical axis has equation (y – k) (x – h) 2 = 1 b a - In this case, transverse axis is vertical. c = a b 2 For both equations. NOTE: These two hyperbolas are graphed with center (0,0)

7 Standard 4, 9, 16, 17 Given the graph below obtain the equation of the hyperbola. 4 2 6 -2 -4 -6 8 10 -8 -10 x y 13 ( , 0) ( , 0) (-1,0) (5,0) Transverse axis is 6 units: 2a=6 a = 9 2 Focus2 Focus1 a = 3 then we know: (2,0) c = a b 2 -a a 2 b = c - a 2 b = 2 13 -9 From the figure: b = 13-9 2 h= 2 Center = (2,0) If k= 0 b = 4 2 Focus1 =( h+c,k) Hyperbola is horizontal: 2+c (x – h) (y – k) 2 = 1 a b - (x-(+2)) (y-(0)) 2 =1 9 4 - = 13 ( , 0) 2+c (x-2) y 2 =1 9 4 - then 2 + c = 2 + 13 c = 13 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

8 Standard 4, 9,16, 17 Given the graph below obtain the equation of the hyperbola. 4 2 6 -2 -4 -6 8 10 -8 -10 x y Transverse axis is 4 units: Focus1 2a=4 13 (4, ) a = 2 a = 4 2 then we know: (4,-2) c = a b 2 -a a 2 13 (4, ) b = c - a 2 b = 2 13 -4 Focus2 From the figure: b = 13-4 2 h= 4 Center = (4,-2) If k= -2 b = 9 2 Focus1 =( h,k+c) Hyperbola is vertical: -2+c (y – k) (x – h) 2 = 1 a b - (y-(-2)) (x-(4)) 2 =1 4 9 - = 13 (4, ) -2+c (y+2) (x-4) 2 =1 4 9 - then -2 + c = -2 + 13 c = 13 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

9 - - Standard 4, 9, 16, 17 Draw the hyperbola that is represented by:
(x-3) (y-4) 2 =1 25 9 - (x -(+3)) (y-(+4)) 2 =1 25 9 - h= 3 Center = (3,4) k= 4 4 2 6 -2 -4 -6 8 10 -8 -10 x y a = 25 2 a = 5 b = 9 2 b= 3 c = a b 2 c = 2 c = 34 2 c= 34 5.8 Focus 1= (h+c, k) = (3+5.8,4) = (8.8,4) Focus 2= (h-c, k) = (3-5.8,4) = (-2.8,4) PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

10 - - Standard 4, 9, 16, 17 Draw the hyperbola that is represented by:
(y-6) (x+2) 2 =1 81 64 - (y-(+6)) (x-(-2)) 2 =1 81 64 - h= -2 Center = (-2,6) k= 6 a = 81 2 a = 9 y 8 4 12 -4 -8 -12 16 20 -16 -20 x b = 64 2 b= 8 c = a b 2 c = 2 c = 145 2 c= 145 12 Focus 1= (h, k+c) = (-2,6+12) = (-2,18) Focus 2= (h, k–c ) = (-2,6-12) = (-2,-6) PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

11 Standard 4, 9,16, 17 = 1 a b 36x - 25y -144x +150y - 981 = 0
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 = 1 a b - 36x y x +150y = 0 2 36x x y y = 0 2 36x -36(4)x y -(-25)6y = 0 2 4 2 6 2 36 x x y y = 2 (2) 2 (3) 2 36 x x y y = 2 36 x x y y = 2 4 (4) 9 (9) 36(x-2) - 25(y-3) = 2 36(x-2) (y-3) = -81 2 36(x-2) (y-3) = 900 2 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

12 Standard 4, 9,16, 17 36x y x +150y = 0 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 = 1 a b - (x-(+2)) (y-(+3)) 2 =1 25 36 - a = 25 2 a = 5 36(x-2) - 25(y-3) = 900 2 b = 36 2 b= 6 (x – h) (y – k) 2 = 1 a b - 36(x-2) (y-3) 2 =1 900 - 36(x-2) (y-3) 2 =1 900 36 25 - h= 2 Center = (2,3) 4 2 6 -2 -4 -6 8 10 -8 -10 x y k= 3 c = a b 2 c = 2 (x-2) (y-3) 2 =1 25 36 - c = 61 2 c= 61 7.8 Focus 1= (h+c, k) = (2+7.8,3) = (9.8,3) Focus 2= (h-c, k) = (2-7.8,3) = (-5.8,3) PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

13 - - + Standard 4, 9,16, 17 Equations for the asymptotes.
First we find the slope for both asymptotes: Center = (2, 3) y 1 3 x 1 2 6 6 + = 6 5 - m= m= 5 - 5 Using the center and the point-slope form of the equation of a line: 4 2 6 -2 -4 -6 8 10 -8 -10 x y (y – ) = m(x – ) y 1 x 1 6 + 6 + 6 5 6 5 - (y – ) = (x – ) (y – ) = (x – ) 3 2 -5 5 + y - 3 = (x – 2) 6 5 y – 3 = (x – 2) 6 5 - y - 3 = x - 12 5 6 y – 3 = x + 12 5 6 - y = x - 12 5 6 15 y = x + 12 5 6 - 15 + 15 5 + 15 5 y = x + 3 5 6 y = x + 27 5 6 - 3 5 = 15 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

14 Conic Section Relationship of A and C
Standard 4, 9, 16, 17 Equation of a Conic Section Ax + Bxy + Cy + Dx + Ex + F = 0 2 Conic Section Relationship of A and C Parabola A=0 or C=0, but not both Circle A=C Ellipse A 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! PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

15 Standard 4, 9, 16, 17 Match the following equations with the corresponding graph: y Ax + Bxy + Cy + Dx + Ex + F = 0 2 3 1) x + y + 8x +2y -32=0 2 A=1 C=1 x y 2) 16x y -64x -384y = 0 2 4 A=+16 C=+64 x y 3) 4x + 24x – y +16 =0 2 2 A=4 C=0 x y 4) 36x y x +150y = 0 2 1 A=+36 C= -25 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

16 Solve the following system of equations:
Standard 4, 9, 16, 17 Solve the following system of equations: y = x – 2x – 3 2 y = -3x – 1 Graphing the line: y = -3x – 1 Graphing the parabola: = -3 +1 m = -3 y = x – 2x – 3 2 b= -1 4 2 6 -2 -4 -6 8 10 -8 -10 x y y = x – 2x + 1 – 3 – 1 2 x= 1 y = -4.25 y = (x – 1) – 4 2 h= 1 ( 1, ) 1 4 (-2, 5) Focus: k= - 4 1 a= 1 = ( 1, 3.75) Vertex: (1, - 4) (1,- 4) Axis of symmetry: x= 1 Latus rectum: 1 The solution is: (-2, 5) and (1, - 4) Now let’s check this result algebraically! y = -4 - 1 4( ) Directrix: = 1 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

17 Solve the following system of equations:
Standard 4, 9, 16, 17 Solve the following system of equations: y = x – 2x – 3 2 4 2 6 -2 -4 -6 8 10 -8 -10 x y y = -3x – 1 x= 1 y = -4.25 Solving by substitution: (-2, 5) -3x – 1 = x – 2x – 3 2 (1,- 4) -3x = x – 2x – 2 2 -2 1 +3x x 0 = x – x – 2 2 (2)(-1) 2 + -1= 1 0 = (x + 2)(x – 1) Using: x= -2 Using: x= 1 x + 2= 0 x – 1= 0 y = -3( ) – 1 -2 y = -3( ) – 1 1 +1 +1 y = 6 – 1 y = -3 – 1 x = -2 x = 1 y = 5 y = -4 (-2, 5) (1, - 4) PRESENTATION CREATED BY SIMON PEREZ. All rights reserved