Report

10.4 - Ellipses Ellipses - Warm Up Solve each equation. 1. 27 = x2 + 11 2. x2 = 48 3. 84 = 120 – x2 Ellipses – Warm Up Solutions 1. 27 = x2 + 11 16 = x2 x = ± 2. x2 = 48 3. x = ± x = ±4 16 = ±4 48 3 84 = 120 – x2 –36 = –x2 36 = x2 x = ± 36 = ±6 ELLIPSE TERMS Vertex EQUATION Co-vertex FORM Focus ( x h)2 ( y k )2 1 a2 b2 CENTER (h, k ) VERTICES (h ± a , k) a V=(h , k) c CO-VERTICES (h, k ± b) b MAJOR AXIS horizontal Co-vertex Vertex MINOR AXIS Major axis Minor axis c a b 2 MAJOR length 2 2 2a vertical MINOR length 2b FOCI (h ± c , k) ELLIPSE TERMS Vertex Minor axis a V=(h , k) Co-vertex b Co-vertex EQUATION FORM ( x h)2 ( y k )2 1 b2 a2 CENTER (h, k ) VERTICES (h, k ± a) c CO-VERTICES (h ± b , k) Focus Vertex MAJOR AXIS MAJOR length Major axis MINOR AXIS MINOR length c a b 2 2 2 FOCI vertical 2a horizontal 2b (h , k ± c ) CONVERTING to STANDARD FORM x² + 4y² + 4x – 24y + 24 = 0 Groups the x terms and y terms x² + 4x + 4y² – 24y + 24 = 0 Complete the square x² + 4x + 4(y² – 6y) + 24 = 0 x² + 4x + 4 + 4(y² – 6y + 9) = -24 + 4 +36 (x + 2)² + 4(y – 3)² = 16 Divide to put in standard form (x + 2)²/16 + 4(y – 3)²/16 = 1 ( x 2)2 ( y 3)2 1 16 4 Writing an Equation of an Ellipse Write an equation in standard form of an ellipse that has a vertex at (0, –4), a co-vertex at (3, 0), and is centered at the origin. Since (0, –4) is a vertex of the ellipse, the other vertex is at (0, 4), and the major axis is vertical. Since (3, 0) is a co-vertex, the other co-vertex is at (–3, 0), and the minor axis is horizontal. So, a = 4, b = 3, a2 = 16, and b2 = 9. x2 b2 + y2 a2 = 1 x2 9 + y2 16 = 1 An equation of the ellipse is Standard form for an equation of an ellipse with a vertical major axis. Substitute 9 for b2 and 16 for a2. x2 y2 9 + 16 = 1. Graph and Label b) Find coordinates of vertices, covertices, foci ( x 3 ) 2 ( y 2) 2 1 9 4 Center = (-3,2) Horizontal ellipse since the a² value is under x terms Since a = 3 and b = 2 Vertices are 3 points left and right from center (-3 ± 3, 2) Covertices are 2 points up and down (-3, 2 ± 2) Now to find focus points Use c² = a² - b² So c² = 9 – 4 = 5 c² = 5 and c = ±√5 Focus points are √5 left and right from the center F(-3 ±√5 , 2) • • • • a) GRAPH Plot Center (-3,2) a = 3 (go left and right) b = 2 (go up and down) Graph and Label b) Find coordinates of vertices, covertices, foci ( x 3)2 ( y 1)2 1 3 16 Center = (3,-1) Vertical ellipse since the a² value is under y terms Since a = 4 and b = 2 Vertices are 3 points up and down from center (3, -1 ± 2) Covertices are 2 points left and right (3 ± 2, -1) Now to find focus points Use c² = a² - b² So c² = 16 – 4 = 12 c² = 12 and c = ±2√3 Focus points are 2√3 up and down from the center F(3,-1 ±2√3) • • • • a) GRAPH Plot Center (3,1) a = 4 (go up and down) b = 2 (go left and right) Write the equation of the ellipse given… endpoints of major axis are at (-11,5) and (7,5) endpoints of minor axis are at (-2,9) and (-2,1) Draw a graph with given info Use given info to get measurements ◦ Major axis = 2a ◦ Major axis is 18 units, ◦ so a = 9 A = (-11,5) ◦ Minor axis = 2b ◦ Minor axis is 8 units, ◦ so b = 4 Use standard form ( x h)2 ( y k )2 1 a2 b2 ◦ Need values for h,k, a and b ◦ We know a = 9 and b = 4 ◦ How do we find center??? ◦ Use midpoint formula ◦ (h , k) = (-2 , 5) Plug into formula ( x 2)2 ( y 5)2 1 92 42 C = (-2,9) minor major D = (-2,1) B = (7,5) Let’s Try One Find an equation of an ellipse centered at the origin that is 20 units wide and 10 units high. Since the largest part of the ellipse is horizontal and the width is 20 units, place the vertices at (±10, 0). Place the co-vertices at (0, ±5). So, a = 10, b = 5, a2 = 100, and b2 = 25. x2 a2 + y2 b2 = 1 x2 y2 100 + 25 = 1 Standard form for an ellipse with a horizontal major axis. Substitute 100 for a2 and 25 for b2. x2 y2 An equation of the ellipse is 100 + 25 = 1. Write the equation of the ellipse given… foci are at (2,5) and (-2,5) vertex at (-3,5) Draw a graph with given info Use given info to get measurements ◦ Find the center first. ◦ The center is in the middle of the foci. Use midpoint formula to find (h , k) ◦ (h , k) = (0 , 5) ◦ Then c = distance from center to foci ◦ So c = 2 ◦ Then a = distance from center to vertex ◦ so a = 3 Use standard form ( x h)2 ( y k )2 1 a2 b2 ◦ Need values for h,k, a and b ◦ We know a = 3, c = 2, (h,k) = (0,5) ◦ How do we find b??? ◦ Use c² = a² – b² ◦ 4 = 9 – b² ◦ we get b² = 5 Plug into formula ( x ) 2 ( y 5 )2 1 9 5 a=3 c=2 (h,k) = (0,5) Working Backwards Find the foci of the ellipse with the equation 9x2 + y2 = 36. Graph the ellipse. 9x2 + y2 = 36 y2 x2 + =1 4 36 Write in standard form. Since 36 > 4 and 36 is with y2, the major axis is vertical, a2 = 36, and b2 = 4. c2 = a2 – b2 = 36 – 4 Find c. Substitute 4 for a2 and 36 for b2. = 32 c=± 32 = ± 4 2 The major axis is vertical, so the coordinates of the foci are (0, ±c). The foci are: (0, 4 2 ) and (0, – 4 2). Let’s Try One Write an equation of the ellipse with foci at (0, ±4) and covertices at (±2, 0). Since the foci have coordinates (0, ±4), the major axis is vertical. Since c = 4 and b = 2, c2 = 16, and b2 = 4. c2 = a2 – b2 Use the equation to find a2. 16 = a2 – 4 Substitute 16 for c2 and 4 for b2. a2 = 20 Simplify. x2 y2 + =1 4 20 An equation of the ellipse is Substitute 20 for a2 and 4 for b2. y2 x2 + 4 20 = 1. Application of foci in ellipses You may think that most objects in space that orbit something else move in circles, but that isn't the case. Although some objects follow circular orbits, most orbits are shaped more like "stretched out" circles or ovals. Mathematicians and astronomers call this oval shape an ellipse. The Sun isn't quite at the center of a planet's elliptical orbit. An ellipse has a point a little bit away from the center called the "focus". The Sun is at the focus of the ellipse. Because the Sun is at the focus, not the center, of the ellipse, the planet moves closer to and further away from the Sun every orbit More on orbits… Orbits are ellipses. An ellipse can be like a circle, or it can be long and skinny. Mathematicians and astronomers use the term "eccentricity" to describe the shape of an orbit. Eccentricity = c/a. An orbit shaped almost like a circle has a low eccentricity close to zero. A long, skinny orbit has a high eccentricity, close to one http://windows2universe.org/physical_science/physics/mechanics/orbit/orbi t_shape_interactive.html