30 Ellipses 1 An ellipse is a set

30. Ellipses 1

An ellipse is a set of points in a plane the sum of whose distances from two fixed points, called foci, is a constant. For any point P that is on the ellipse , d 2 + d 1 is always the same. P d 1 F 1 d 2 F 2

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(x – h)2 + (y – k)2 = 1 a 2 b 2 The center of the ellipse is at the point (h, k) a is ½ the length of the horizontal axis b is ½ the length of the vertical axis Where c is the distance from the center to a focus point. c 2 = a 2 – b 2 if a 2 > b 2 c 2 = b 2 – a 2 if b 2 > a 2

Graph: x 2 + y 2 = 1 4 9 Center: (0, 0) V Minor Axis: 4 (horizontal) F Major Axis: 6 (vertical) Vertices: (0, 3) and (0, - 3) c 2 = 9 – 4 = 5 c = Ö 5 = 2. 24 Foci: (0, 2. 24) and (0, -2. 24) F V

Graph: (x – 2)2 + (y + 1)2 = 1 16 9 Center: (2, -1) Major Axis: 8 (horizontal) Minor Axis: 6 (vertical) V Vertices: (6, -1) and (-2, -1) c 2 = 16 – 9 = 7 c = Ö 7 = 2. 65 Foci: (4. 65, -1) and (-0. 65, -1) V

Find the equation of the ellipse given that Vertices are (0, 4), (0, -4) Foci are (0, 3), (0, -3) V Center: (0, 0) Major Axis: 8 (vertical) b = 4 and b 2 = 16 Since c = 3, and c 2 = b 2 – a 2 9 = 16 – a 2 = 7 Equation: x 2 + y 2 =7 1 9 V (x – h)2 + (y – k)2 = 1 a 2 b 2

Find the equation of the ellipse given the graph Then locate the foci of the ellipse Center: (-1, 1) Major Axis: 6 (horizontal), so a = 3 Minor Axis: 2 (vertical), so b = 1 V 1 Equation: (x + 1)2 + (y – 1)2 = 1 9 1 c 2 = a 2 – b 2 c 2 = 9 – 1 = 8 c = Ö 8 = 2. 83 Foci are (1. 83, 1), (-3. 83, 1) V 2 (x – h)2 + (y – k)2 = 1 a 2 b 2

Find the equation of the ellipse given that Foci are (-2, 0), (2, 0) y-intercepts: -3, 3 Major Axis must be horizontal since the foci are on the major axis Center: (0, 0) c= 2 and b = 3 F F c 2 = a 2 – b 2 4 = a 2 – 9 a 2 = 13 Equation: x 2 + y 2 =131 9 (x – h)2 + (y – k)2 = 1 a 2 b 2

Convert the following equation to standard form Then graph the ellipse and locate its foci 9 x 2 + 25 y 2 – 36 x + 50 y – 164 = 0 9(x 2 – 4 x + ___ 4 ) + 25(y 2 + 2 y + ___) 1 = 164 + 36 ___ + ___ 25 9(x – 2)2 + 25(y + 1)2 = 225 (x – 2)2 (y + 1)2 + = 1 25 9 c 2 = 25 – 9 = 16 c=4 Foci: (-2, -1) and (6, -1) F F

Convert the following equation to standard form Then graph the ellipse and locate its foci 4 x 2 + 9 y 2 – 48 x + 72 y + 144 = 0. 4(x 2 – 12 x + ___ 36 ) + 9(y 2 + 8 y + ___) 16 = -144 +144 ___ + ___ 144 4(x – 6)2 + 9(y + 4)2 = 144 (x – 6)2 (y + 4)2 + = 1 36 9 c 2 = 36 – 9 = 27 c = 5. 2 Foci: (. 8, -4) and (11. 2, -4) F F

A semielliptical archway has a height of 20 feet at its midpoint and a width of 50 feet. Can a truck that is 14 ft high and 10 ft wide drive under the archway without moving into the oncoming lane? The real question is “What is the value of y at point P when x = 10” ? Equation of the ellipse: x 2 + y 2 = 1 625 400 P 20 10 50 16 x 2 + 25 y 2 = 10, 000 When x = 10, 1600 + 25 y 2 = 10, 000 25 y 2 = 8400 y 2 = 336 y = 18. 3 Yes, the truck will be able to drive under the archway without moving into the oncoming lane.