1. 5.4 Hyperbolas 1 Please note the minus in the middle. A “+” in the middle makes the graph an ellipse. A minus in the middle will give us a hyperbola which looks like the following graphs below. In the last section, we graphed ellipses of the form: We will now “focus” on equations of the form: and Next Slide ●● ● ●

2. 5.4 Hyperbolas 2 Next Slide Definition: A hyperbola is the collection of all points in the plane the difference of whose distance from two fixed points called the foci is a positive constant. ●● ● ● ● ● ● ● (c,0) (-c,0) (a,0) (-a,0) (0,b) (0,-b) foci (0,c) (0,-c) vertices Horizontal Transverse Axis Vertical Transverse Axis A hyperbola which opens left and right has a horizontal transverse axis. A hyperbola which opens up and down has a vertical transverse axis. The midpoint of the transverse axis is the center.

3. 5.4 Hyperbolas 3 Standard Equation: Hyperbola with Transverse Axis on the x Axis (Opens Left and Right) The standard equation of a hyperbola with its center at (0,0) and its transverse axis on the x axis is where the vertices are (-a,0) and (a,0), and the foci are at (-c,0) and (c,0), and The hyperbola’s shape will also be determined by two asymptotes. The equation of these asymptotes are the lines: Some instructors will require you to state the equations of these asymptotes and other instructors may not. The equations of the asymptotes are simple to state when the center is at the origin. If the center is not at the origin, the equation of the asymptotes are: Next slide

4. 5.4 Hyperbolas 4 x y The vertices are (-3,0) and (3,0). Next Slide Plot all four of these points (-3,0), (3,0), (0,2) and (0,-2). Now, create a dashed box using the four points. Next, create the two dashed asymptotes drawing diagonals through the corners of the box. The two branches of the hyperbola opening left and right at the vertices can be sketched using the diagonals as guidelines. The branches get closer and closer to the asymptotes but never cross it.

5. 5.4 Hyperbolas 5 Your Turn Problem #1 y vertices: (-2,0), (2,0) Answer x

6. 5.4 Hyperbolas 6 Next slide x is positive: opens left and right y is positive: opens up and down Standard Equation: Hyperbola with Transverse Axis on the y Axis (Opens up and down) The standard equation of a hyperbola with its center at (0,0) and its transverse axis on the y axis is where the vertices are (0,-b) and (0,b), and the foci are at (0,-c) and (0,c), and The hyperbola’s shape will again be determined by two asymptotes. The equation of these asymptotes are still the lines: It may seem confusing to determine if the branches open left and right or up and down. Once the hyperbola is in standard form, the vertices from which the branches open will be on the axis that comes first in the equation, which will have a positive coefficient.

7. 5.4 Hyperbolas 7 x y The vertices are (0,-2) and (0,2). Next Slide Plot all four of these points (-3,0), (3,0), (0,2) and (0,-2). Now, create a dashed box using the four points. Next, create the two dashed asymptotes by drawing diagonals through the corners of the box. The two branches of the hyperbola opening up and down at the vertices can be sketched using the diagonals as guidelines.

8. 5.4 Hyperbolas 8 Your Turn Problem #2 x y vertices: (0,-4), (0,4) Answer:

9. 5.4 Hyperbolas 9 4 44 x y The vertices are (-2,0) and (2,0). Next Slide Now, create a dashed box using the four points. Next, create the two dashed asymptotes by drawing diagonals through the corners of the box. The two branches of the hyperbola opening left and right at the vertices can be sketched using the diagonals as guidelines. Plot all four of these points (0,-1), (0,1), (2,0) and (-2,0). To obtain the standard form, divide both sides by 4 and simplify to obtain a “1” on the RHS.

10. 5.4 Hyperbolas 10 Your Turn Problem #3 x y vertices: (0,-5), (0,5) Answer:

11. 5.4 Hyperbolas 11 Hyperbolas whose center is not at the origin. The standard form for a hyperbola where the center is at (h, k) is: This hyperbola opens left and right. Next Slide or This hyperbola opens up and down. y x (h,k) (h,k+b) (h,k-b) ● ● ● ● ● (h-a,k) ● (h+a,k) (h,k+c) (h,k-c) ● x (h-a,k)(h-c,k) ● ● (h+c,k) ● (h+a,k) ● (h,k+b) (h,k-b) ● ● y ● (h,k)

12. 5.4 Hyperbolas 12 x y The vertices are (2,1) and (2,-7). (2,-3) (2,-7) (2,1) (0,-3) (4,-3) Next Slide Plot all four of these points (0,-3), (4,-3), (2,1) and (2-7). Now, create the dashed box using the four points and draw the asymptotes through the corners. Then sketch the hyperbola. Compare to the standard form to determine the center, a, and b.

13. 5.4 Hyperbolas 13 Your Turn Problem #4 (-1,-1) (-1,2) (-6,2) x ●● ● ● center: (-1,2) vertices: (-6,2), (4,2) Answer: (4,2) (-1,5) ● ● y ●

14. 5.4 Hyperbolas 14 The hyperbola in the previous example was given in standard form. If the hyperbola is given in general form, we will need to convert it to standard form before graphing. Next Slide or Factoring and completing the square will be necessary to obtain the standard form:

15. 5.4 Hyperbolas 15 Example 5.Write the given equation of the hyperbola in standard form: Group the x terms separately from the y terms and move the constant to the RHS. Complete the square for both trinomials. The numbers added in the parentheses are 1 and 9. We need to add the same “value” to the RHS. The values are 4 and -81. 19 Write each perfect square trinomial as a binomial squared and add the constants on the RHS. Factor out the ‘4” from the first pair and the ‘-9’ from the second pair. Leave a space at the end of each set of parentheses to add the appropriate number when completing the square. Finally, divide by 36 on both sides and simplify to obtain the hyperbola in standard form. 36 Answer: Your Turn Problem #5 Write the given equation of the hyperbola in standard form:

16. 5.4 Hyperbolas 16 x y The vertices are (-1,1) and (5,1). 1 st write in standard form using completing the square. (5,1) (-1,1) Next Slide +8 - 9 (2,1)

17. 5.4 Hyperbolas 17 (-3,-1) (-1,1)(-5,1) x ● ● ● ●● y center: (-3,1) vertices: (-3,3), (-3,-1) Answer: (-3,3) ● ● (-3,1) The End B.R. 2-1-07 Your Turn Problem #6