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C O N I C S E C T I O N S Part 3: The Ellipse. Circle Ellipse Foci Vertex Co-vertex The Major Axis is the longest segment that cuts the ellipse in half.

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Presentation on theme: "C O N I C S E C T I O N S Part 3: The Ellipse. Circle Ellipse Foci Vertex Co-vertex The Major Axis is the longest segment that cuts the ellipse in half."— Presentation transcript:

1 C O N I C S E C T I O N S Part 3: The Ellipse

2 Circle Ellipse Foci Vertex Co-vertex The Major Axis is the longest segment that cuts the ellipse in half. The Minor Axis is the shortest segment that cuts the ellipse in half. Vertex It intersects with the ellipse at the Vertices. It intersects with the ellipse at the Co-vertices. Co-vertex An ellipse has 2 Focus points that are on the Major Axis and equidistant from the Center of the ellipse. Center

3 Standard Equation of an Ellipse (x-h) 2 + (y-k) 2 = 1 a 2 b 2 When major axis is horizontal. a = distance from center to vertex b = distance from center to co-vertex c = distance from center to focus c2 = a2 – b2c2 = a2 – b2 (c, 0) (–c, 0) (–a, 0) (a, 0) (0, b) (0, –b)

4 Standard Equation of an Ellipse When major axis is vertical. (x-h) 2 + (y-k) 2 = 1 b 2 a 2 a = distance from center to vertex b = distance from center to co-vertex c = distance from center to focus c2 = a2 – b2c2 = a2 – b2 (0, c) (0, –c) (0, –a) (0, a) (b, 0) (–b, 0)

5 What is the relationship of the denominators? (x-h) 2 + (y-k) 2 = 1 a 2 b 2 a = distance from center to vertex b = distance from center to co-vertex c = distance from center to focus (c, 0) (–c, 0) (–a, 0) (a, 0) (0, b) (0, –b) c2 = a2 – b2c2 = a2 – b2 (0, c) (0, –c) (0, –a) (0, a) (b, 0) (–b, 0) (x-h) 2 + (y-k) 2 = 1 b 2 a 2 Notice that when the major axis is parallel with the x-axis, a 2 goes with the (x-h) 2 ; but when the minor axis is parallel with the x-axis, b 2 goes with the (x-h) 2

6 Mr. Cool Ice Thinks This Stuff is Cool!

7 Write an equation of the ellipse with vertices (0, –3) & (0, 3) and co-vertices (–2, 0) & (2, 0). (x-h) 2 + (y-k) 2 = 1 b 2 a 2 c 2 = a 2 – b 2 to find c. c 2 = 3 2 – 2 2 c 2 = 9 – 4 = 5 c = (0, c) (0, –c) (0, –3) (0, 3) (2, 0)(–2, 0) Since a = 3 & b = 2 The equation is (x-0) 2 + (y-0) 2 = 1 4 9 Let’s Find the Foci So the Foci are at:

8 Example: Write 9x 2 + 16y 2 = 144 in standard form. Find the foci and vertices. 9x 2 + 16y 2 = 144 144 144 144 Use c 2 = a 2 – b 2 to find c. c 2 = 4 2 – 3 2 c 2 = 16 – 9 = 7 c = (c, 0)(–c,0) (–4,0)(4, 0) (0, 3) (0,-3) That means a = 4 b = 3Vertices: Foci: Simplify... x 2 + y 2 = 1 16 9

9 horizontal Center: (2, –3) a = 5, b = 3 Graph (x – 2) 2 + (y + 3) 2 = 1 25 9 (2, 0) (2,–6) (–3,–3)(7, –3) Start at the center 5 units left and right 3 units up and down

10 Find center, vertices and foci for the ellipse 36x 2 + y 2 – 144x + 8y = –124 36(x – 2) 2 + (y + 4) 2 = 36 Group the x’s and y’s together... 36x 2 – 144x + y 2 + 8y = –124 Factor to make the leading coefficients 1 36(x 2 – 4x ) + (y 2 + 8y ) = –124 Complete the squares. + 4 +16 + (36)(4) Set equal to 1 36(x – 2) 2 + (y + 4) 2 = 36 36 36 36 (x – 2) 2 + (y + 4) 2 = 1 1 36 Center: ( 2, – 4 ) Since the major axis is vertical, the vertices will be a units above and below the center. Vertices: ( 2, 2 ) & (2, -10 ) + 16 The foci are c units from the center and c 2 = a 2 – b 2 c 2 = 36 – 1 c 2 = 35 c = a = 6 ; b = 1 Co-vertices: ( 3, - 4 ) & ( 1, - 4 ) Foci: ( 2, - 4 - ) & ( 2, - 4 + )


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