8.5 Graph and Write Equations of Hyperbolas

1. 8.5 Graph and Write Equations of Hyperbolas p.518 What are the parts of a hyperbola? What are the standard form equations of a hyperbola? How do you know which way it opens? Given a & b, how do you find the value of c? How do you graph a hyperbola? Why does drawing a box make graphing easier? How do you write the equation from a graph?

2. Hyperbolas • Like an ellipse but instead of the sum of distances it is the difference • A hyperbola is the set of all points P such that the differences from P to two fixed points, called foci, is constant • The line thru the foci intersects the hyperbola @ two points (the vertices) • The line segment joining the vertices is the transverse axis, and it’s midpoint is the center of the hyperbola. • Has 2 branches and 2 asymptotes • The asymptotes contain the diagonals of a rectangle centered at the hyperbolas center

3. Asymptotes (0,b) Vertex (a,0) Vertex (-a,0) Focus Focus (0,-b)

4. Vertical transverse axis

5. Standard Form of Hyperbola w/ center @ origin Foci lie on transverse axis, c units from the center c2 = a2+b2

6. Graph 4x2 – 9y2 = 36 • Write in standard form (divide through by 36) • a=3 b=2 – because x2 term is ‘+’ transverse axis is horizontal & vertices are (-3,0) & (3,0) • Draw a rectangle centered at the origin. • Draw asymptotes. • Draw hyperbola.

8. 100 y2 25y2 4x2 100 29. soc = 29. 100 4 100 = – The foci are at( 0, + ) (0, + 5.4). x225 – = 1 Graph 25y2 – 4x2 = 100. Identify the vertices, foci, and asymptotes of the hyperbola. SOLUTION STEP 1 Rewrite the equation in standard form. 25y2 – 4x2 = 100 Write original equation. Divide each side by 100. Simplify. STEP 2 Identify the vertices, foci, and asymptotes. Note that a2 = 4 and b2 = 25, so a = 2 and b = 5. The y2 - term is positive, so the transverse axis is vertical and the vertices are at (0, +2). Find the foci. c2= a2+ b2= 22+52= 29.

9. ab 25 + + The asymptotes are y = x x y = or STEP 3 Draw the hyperbola. First draw a rectangle centered at the origin that is 2a = 4units high and 2b = 10units wide. The asymptotes pass through opposite corners of the rectangle. Then, draw the hyperbola passing through the vertices and approaching the asymptotes.

10. x2 x2 65 , 0 65. soc = + The equation is in standard form. 16 16 = 1 The foci are at( + ) y2 49 y2 49 – – Graph the equation. Identify the vertices, foci, and asymptotes of the hyperbola. 1. = 1 SOLUTION STEP 1 STEP 2 Identify the vertices, foci, and asymptotes. Note that a2 = 16 and b2 = 49, so a = 4 and b = 7. The x2 - term is positive, so the transverse axis is horizontal and the vertices are at (+4, 0). Find the foci. c2= a2+ b2= 42+ 72= 65.

11. STEP 3 Draw the hyperbola. First draw a rectangle centered at the origin that is 2a = 8units high and 2b = 14units wide. The asymptotes pass through opposite corners of the rectangle. Then, draw the hyperbola passing through the vertices and approaching the asymptotes.

12. Write the equation of a hyperbola with foci (0,-3) & (0,3) and vertices (0,-2) & (0,2). • Vertical because foci & vertices lie on the y-axis • Center @ origin because focus & vertices are equidistant from the origin • Since c=3 & a=2, c2 = b2 + a2 • 9 = b2 + 4 • 5 = b2 • +/-√5 = b

13. Write an equation of the hyperbola with foci at (– 4, 0) and (4, 0) and vertices at (– 3, 0) and (3, 0). SOLUTION The foci and vertices lie on the x-axis equidistant from the origin, so the transverse axis is horizontal and the center is the origin. The foci are each 4 units from the center, so c = 4. The vertices are each 3 units from the center, soa = 3.

14. = 1 x2 x2 32 9 = 1 y2 7 y2 7 – – Becausec2 = a2 + b2, you haveb2 = c2 – a2. Findb2. b2= c2–a2= 42–32= 7 Because the transverse axis is horizontal, the standard form of the equation is as follows: Substitute 3 for aand 7 for b2. Simplify

15. 4.Foci:(– 3, 0), (3, 0) Vertices:(– 1, 0), (1, 0) = 1 x2 y2 8 12 – = 1 x2 y2 8 – Write an equation of the hyperbola with the given foci and vertices. SOLUTION The foci and vertices lie on the x-axis equidistant from the origin, so the transverse axis is horizontal and the center is the origin. The foci are each 3 units from the center, soc = 3. The vertices are each 1 units from the center, soa = 1. Becausec2 = a2 + b2, you haveb2 = c2 – a2. Findb2. b2= c2–a2= 32–12= 8 Because the transverse axis is horizontal, the standard form of the equation is as follows: Substitute 1 for aand 8 for b2. Simplify

16. 5.Foci:(0, – 10), (0, 10) Vertices:(0, – 6), (0, 6) = 1 = 1 y2 y2 36 362 x2 64 x2 64 – – The foci and vertices lie on the y-axis equidistant from the origin, so the transverse axis is vertical and the center is the origin. The foci are each 10 units from the center, soc = 10. The vertices are each 6 units from the center, soa = 6. SOLUTION Becausec2 = a2 + b2, you haveb2 = c2 – a2. Findb2. b2= c2–a2= 102–62= 64 Because the transverse axis is horizontal, the standard form of the equation is as follows: Substitute 6 for aand 64 for b2. Simplify

17. You can take panoramic photographs using a hyperbolic mirror. Light rays heading toward the focus behind the mirror are reflected to a camera positioned at the other focus as shown. After a photograph is taken, computers can “unwrap” the distorted image into a 360° view. Photography • Write an equation for the cross section of the mirror. • The mirror is 6 centimeters wide. How tall is it?

18. – – x2 5.50 32 5.50 y2 7.90 y2 2.812 x2 5.50 y2 7.90 – = 1 y2 20.83 b2 = c2 – a2 = 3.662 – 2.812 5.50 y – 4.56 = 1 = 1 SOLUTION From the diagram, a= 2.81andc= 3.66. STEP 1 To write an equation, find b2. Because the transverse axis is vertical, the standard form of the equation for the cross section of the mirror is as follows: or STEP 2 Find the y-coordinate at the mirror’s bottom edge. Because the mirror is 6 centimeters wide, substitute x = 3 into the equation and solve. Substitute 3 for x. Solve fory2. Solve fory. So, the mirror has a height of– 2.81 – (– 4.56) = 1.75 cm.

19. What are the parts of a hyperbola? Vertices, foci, center, transverse axis & asymptotes • What are the standard form equations of a hyperbola? • How do you know which way it opens? Transverse axis is always over a • Given a & b, how do you find the value of c? c2 = a2 + b2 • How do you graph a hyperbola? Plot a and b, draw a box with diagonals. Draw the hyperbola following the diagonals through the vertices.

20. Why does drawing a box make graphing easier? The diagonals of the box are the asymptotes of the hyperbola. • How do you write the equation from a graph? Identify the transverse axis, find the value of a and b (may have to use c2 = a2 + b2) and substitute into the equation.

21. 8.5 Assignment, day 1 Page 521, 3-13 odd, 19-25 odd

22. 8.5 Hyperbolas, day 2 What are the standard form equations of a hyperbola if the center has been translated? How do you graph a translated hyperbola? How do you write the equation of a translated hyperbola?

23. Translated Hyperbolas In the following equations the point (h,k) is the center of the hyperbola. Horizontal axis Vertical axis Remember c2 = a2 + b2

24. (x + 1)2 Graph (y + 1)2– = 1. 4 (–1, 0) (–1, –1) (–1, –2) Graphing the Equation of a Translated Hyperbola SOLUTION The y2-term is positive, so thetransverse axis is vertical. Sincea2 = 1 and b2 = 4, you know thata = 1 and b = 2. Plot the center at (h, k) = (–1, –1).Plot the vertices 1 unit above and below the center at (–1, 0)and(–1, –2). Draw a rectangle that is centered at (–1, –1)and is 2a = 2 units high and 2b = 4 units wide.

25. (x + 1)2 Graph (y + 1)2– = 1. 4 (–1, 0) (–1, –1) (–1, –2) Graphing the Equation of a Translated Hyperbola SOLUTION The y2-term is positive, so thetransverse axis is vertical. Sincea2 = 1 and b2 = 4, you know thata = 1 and b = 2. Draw the asymptotes through the corners of the rectangle. Draw the hyperbola so that it passes through the vertices and approaches the asymptotes.

26. (y – 3)2 (x + 1)2 Plot the center, vertices, and foci. The vertices lie a = 2 units above and below the center, at (21, 5) and (21, 1). Because c2 = a2 + b2 = 13, the foci lie c = 13 3.6 units above and below the center, at (– 1, 6.6) and (– 1, – 0.6). 4 9 Graph – = 1 SOLUTION Compare the given equation to the standard forms of equations of hyperbolas. The equation’s form tells you that the graph is a hyperbola with a vertical transverse axis. The center is at (h, k) = (– 1, 3). Because a2 = 4 and b2 = 9, you know that a = 2 and b = 3. STEP 1 STEP 2

27. STEP 3 Draw the hyperbola. Draw a rectangle centered at (−1, 3) that is 2a = 4 units high and 2b = 6 units wide. Draw the asymptotes through the opposite corners of the rectangle. Then draw the hyperbola passing through the vertices and approaching the asymptotes.

28. (y – 4)2 Plot the center, vertices, and foci. The vertices lie a = 1 units above and below the center, at (–2, 4) and (–4, 4). Because c2 = a2 + b2 = 5, the foci lie c = 5 units above and below the center, at (– 3 + ,4 ) and (– 3, – , 4). 9 5 5 3. Graph (x + 3)2 – = 1 SOLUTION STEP 1 Compare the given equation to the standard forms of equations of hyperbolas. The equation’s form tells you that the graph is a hyperbola with a vertical transverse axis. The center is at (h, k) = (– 3, 4). Because a2 = 1 and b2 = 4, you know that a = 1 and b = 2. STEP 2

29. STEP 3 Draw the hyperbola. Draw a rectangle centered at ( –3, 4) that is 2a = 2 units high and 2b = 4 units wide. Draw the asymptotes through the opposite corners of the rectangle.

30. Write the equation of the hyperbola in standard form. 16y2 −36x2 + 9 = 0 16y2 −36x2 = −9

31. Write an equation for the hyperbola. Vertices at (5, −4)and(5,4)and foci at(5,−6)and(5,6). Draw a quick graph. Equation will be:

32. (5,6) (5,4) Center(5,0)(h,k) a = 4, c = 6 c2 = a2 + b2 62 = 42 + b2 36 = 16 + b2 20= b2 Center? (5,−4) (5,−6)

33. What are the standard form equations of a hyperbola if the center has been translated?

34. 8.5 Assignment, day 2 Page 521, 4-10 even, 18-24 even Page 531, 5, 6, 11, 19-20