10-5 Hyperbolas Hubarth Algebra II

hyperbola. The midpoint of the segment is the center of the hyperbola. A hyperbola is a set of points 𝑃 in a plane such that the absolute value of the difference between the distances from 𝑃 to two fixed points 𝐹 1 and 𝐹 2 is a constant 𝑘. 𝑃 𝐹 1 −𝑃 𝐹 2 =𝑘, where 𝑘< 𝐹 1 𝐹 2 Each fixed point F is a focus of a hyperbola. The segment that lies on the line containing the foci and the endpoints on a hyperbola is the transverse axis. The endpoints are the vertices of a hyperbola. The midpoint of the segment is the center of the hyperbola. 𝑥 2 𝑎 2 − 𝑦 2 𝑏 2 =1 Standard form of an equation of a hyperbola with a horizontal transverse axis

𝑥 2 𝑎 2 − 𝑦 2 𝑏 2 =1 Standard form of an equation of a hyperbola with a horizontal transverse axis 𝒚= 𝒃 𝒂 𝒙 𝒚=− 𝒃 𝒂 𝒙 foci: 𝐹 1 , 𝐹 2 vertices: ±𝑎, 0 asymptotes: 𝑦=± 𝑏 𝑎 𝑥 x-intercept: ±𝑎 y-intercept: none . . (-a, 0) (a, 0) 𝑭 𝟏 𝑭 𝟐 Transverse Axis 𝑦 2 𝑎 2 − 𝑥 2 𝑏 2 =1 Standard form of an equation of a hyperbola with a vertical transverse axis foci: 𝐹 1 , 𝐹 2 vertices: (0, ±𝑎) asymptotes: 𝑦=± 𝑏 𝑎 𝑥 x-intercept: none y-intercept: ±𝑎 . 𝑭 𝟏 𝒚=− 𝒃 𝒂 𝒙 𝒚= 𝒃 𝒂 𝒙 (0, a) . (0, -a) 𝑭 𝟐 Transverse Axis

Ex. 1 Graphing Hyperbolas Graph 4x2 – 16y2 = 64. The equation of the form 𝑥 2 𝑎 2 − 𝑦 2 𝑏 2 =1, so the transverse axis is horizontal. 4x2 – 16y2 = 64 – = 1 x 2 16 y 2 4 Since a2 = 16 and b2 = 4, a = 4 and b = 2. Step 1: Graph the vertices. Since the transverse axis is horizontal, the vertices lie on the x-axis. The coordinates are (±a, 0), or (±4, 0). Step 2: Use the values a and b to draw the central rectangle. The lengths of its sides are 2a and 2b, or 8 and 4. Step 3: Draw the asymptotes. The equations of the asymptotes are 𝑦=± 𝑏 𝑎 or 𝑦=± 1 2 𝑥. The asymptotes contain the diagonals of the central rectangles Step 4: Sketch the branches of the hyperbola through the vertices so they approach the asymptotes. Quick Check

Ex. 2 Finding the Foci of a Hyperbola Find the foci of the graph 𝑦 2 4 − 𝑥 2 9 =1. Draw the graph The equation is in the form – = 1, so the transverse axis is horizontal; a2 = 9 and b2 = 4. x 2 a2 y 2 b2 c2 = a2 + b2 Use the Pythagorean Theorem. = 9 + 4 Substitute 9 for a2 and 4 for b2. c = 13 3.6 Find the square root of each side of the equation. The foci (0, ±c) are approximately (0, –3.6) and (0, 3.6). The vertices (0, ±b) are (0, –2) and (0, 2). The asymptotes are the lines y = ± x , or y = ± x. b a 2 3

Ex. 3 Real-World Connection As a spacecraft approaches a planet, the gravitational pull of the planet changes the spacecraft’s path to a hyperbola that diverges from its asymptote. Find an equation that models the path of the spacecraft around the planet given that a = 300,765 km and c = 424,650 km. Assume that the center of the hyperbola is at the origin and that the transverse axis is horizontal. The equation will be in the form – = 1. x 2 a2 y 2 b2 c2 = a2 + b2 Use the Pythagorean Theorem. (424,650)2 = (300,765)2 + b2 Substitute. 1.803  1011 = 9.046  1010 + b2 Use a calculator. b2 = 1.803  1011 – 9.046  1010 Solve for b2. = 8.987  1010 The path of the spacecraft around the planet can be modeled by 𝑥 2 9.046 ∗ 10 10 − 𝑦 2 8.987∗ 10 10 =1

1. Graph the hyperbola with equation 𝑦 2 16 − 𝑥 2 9 =1 Practice 1. Graph the hyperbola with equation 𝑦 2 16 − 𝑥 2 9 =1 2. Find the foci of 𝑥 2 25 − 𝑦 2 9 =1. Draw the graph. 34 , 0 , (− 34 , 0) 3. Find an equation that models the path of Voyager 2 around Jupiter, given that a = 2, 184,140 km and c = 2, 904, 906.2 km 𝑥 2 4.777∗ 10 12 − 𝑦 2 3.668∗ 10 12 =1