Presentation is loading. Please wait.

Presentation is loading. Please wait.

Conics Hyperbola. Conics Hyperbola Cross Section.

Similar presentations

Presentation on theme: "Conics Hyperbola. Conics Hyperbola Cross Section."— Presentation transcript:

1 Conics Hyperbola

2 Conics

3 Hyperbola Cross Section

4 Hyperbola A hyperbola is the set of all points in a plane whose distances from two fixed points in the plane have a constant difference. The fixed points are the foci of the hyperbola. The line through the foci is the focal (major) axis. The point on the focal axis midway between the foci is the center. The points where the hyperbola intersects its focal axis are the vertices of the hyperbola.

5 Horizontal

6 Vertical

7 Axis The chord lying on the focal axis connecting the vertices is the transverse axis. Semi transverse axis—the distance from the center to the vertex—(a)

8 Hyperbola with center (0,0) HorizontalVertical STD Focal Axisx-axisy-axis Foci (±c,0)(0, ±c) Vertices(±a,0)(0, ±a) Semi-Tran Axis a a Semi-Conjugate Axis b b Pythagorean Rel. c²=a²+b² c²=a²+b² Asymptotes y=±(b/a)x y=±(a/b)x

9 E1 1) Plot x² - y² = 1

10 E1 a 2 = 1, so a = 1 b 2 = 1, so b = 1 This equation is for a hyperbola whose center is at the origin. So sketch in the green square. Draw the green lines through the diagonals of the square, these are the asymptotes. The vertices occur at y=0, substituting into the equation we get: x 2 - 0 = 1. x = ± 1 Plot the vertices (red dots) and sketch the branches without crossing the asymptotes.

11 E2 Plot (x²/9) – (y²/16) = 1

12 E2 Here a 2 = 9, so a = 3 and b 2 = 16, so b = 4 when y = 0, x 2 = 9 so the vertices are at x = ± 3 Plot the green rectangle, sketch in the asymptotes, and mark the vertices. Now sketch in the hyperbola without crossing the asymptotes.

13 Hyperbola with Center (h,k)

14 E3 Plot (y²/49) – (x²/4) = 1

15 E3 Notice that the signs have interchanged, the minus is in front of x 2 and the plus sign is in front of y 2. This is a hyperbola the opens along the Y axis. We have b 2 = 49, so b = 7 and a 2 = 4, so a = 2. The vertices are at x=0, substituting in we get y 2 / 49 - 0 = 1 which is y 2 = 49 so y = ±7 Plot the green rectangle, the asymptotes through its diagonals and the vertices then sketch in the hyperbola without crossing the asymptotes.

Download ppt "Conics Hyperbola. Conics Hyperbola Cross Section."

Similar presentations

Ads by Google