5.3 Ellipses 1 We define an ellipse as the set of points which are a fixed distance from two points (the foci), i.e. that the sum of the two distances from any point on the ellipse to the two foci is the same no matter where you are on the ellipse. An ellipse can be constructed using a piece of string and two thumbtacks. Fix the two ends of the string so the string is not tight (very loose). Then with a pencil pull the string so that the string is tight and move the string around to form the ellipse. ● ● P F'F'F A line that passes through both foci and intersects the ellipse at two points (the vertices) is known as the major axis. The minor axis is a chord that is perpendicular to the major axis. Their point of intersection is the center. The major axis, containing the foci, is always longer than the minor axis. vertex center Horizontal Major Axis focus ● ●● ● ● center focus vertex Vertical Major Axis ● ● ● ● ●

5.3 Ellipses 2 Standard Equation: Ellipse with Major Axis on the x Axis The standard equation of an ellipse with its center at (0,0) and its major axis on the x axis is where a > b. The vertices are (  a,0) and (a,0), and the length of the major axis is 2a. The endpoints of the minor axis are (0,  b) and (0,b), and the length of the minor axis is 2b. (-c,0) ● ● (c,0) ● (-a,0) (a,0) (0,b) (0,-b) ● ● ● ● Next slide

5.3 Ellipses 3 To obtain the standard form, divide both sides by 36 and simplify to obtain a “1” on the RHS. 36 x y The vertices (endpoints of the major axis) are (-3,0) and (3,0). The endpoints of the minor axis are (0,2) and (0,-2). Next Slide

5.3 Ellipses 4 Your Turn Problem #1 x y vertices: (-5,0), (5,0) endpoints: (0,-2), (0,2) Answer

5.3 Ellipses 5 Standard Equation: Ellipse with Major Axis on the y Axis The standard equation of an ellipse with its center at (0,0) and its major axis on the y axis is where b > a. The vertices are (0,-b) and (0,b), and the length of the major axis is 2b. The endpoints of the minor axis are (-a,0) and (a,0), and the length of the minor axis is 2a. (0,b) (0,-c) ● (0, c) ● (-a,0) (a,0) (0,-b) ● ● ● ● Next slide

5.3 Ellipses 6 To obtain the standard form, divide both sides by 225 and simplify to obtain a “1” on the RHS. 225 The vertices (endpoints of the major axis) are (0,-5) and (0,5). The endpoints of the minor axis are (-3,0) and (3,0). Next Slide x y

5.3 Ellipses 7 Your Turn Problem #2 x y vertices: (0,-3), (0,3) endpoints: (-1,0), (1,0) Answer

5.3 Ellipses 8 Ellipses whose center is not at the origin. The standard form for an ellipse where the center is not at the origin is where the center is (h,k). If a>b, then the ellipse has a horizontal major axis. If a<b, then the ellipse has a vertical major axis. Also, the foci which lie on the major axis will be a distance of ‘c’ units from the center. y x (h,k) (h,k+b) (h,k-b) ● ● ● ● ● (h-a,k) ● (h+a,k) (h,k+c) (h,k-c) ● x (h-a,k)(h-c,k) ● (h,k) ● ● (h+c,k) ● (h+a,k) ● (h,k+b) (h,k-b) ● ● y Next Slide

5.3 Ellipses 9 x y The vertices (endpoints of the major axis) are (2,1) and (2,-7). The endpoints of the minor axis are (0,-3) and (4,-3). (2,-3) (2,-7) (2,1) Next Slide (0,-3) (4,-3)

5.3 Ellipses 10 (-1,-1) (-1,2) (-6,2) x ●● ● ● ● y center: (-1,2) vertices: (-6,2), (4,2) endpoints: (-1,5), (-1,-1) foci: (-5,2), (3,2) Answer: (4,2) (-1,5) ● (3,2) ● (-5,2) Your Turn Problem #3

5.3 Ellipses 11 Procedure:Writing an ellipse in standard form given the general form. 1.Move the constant to the right hand side and rearrange the terms as follows: (ax 2 + cx + __) + (by 2 + dy + ___) = -e. The ellipse in the previous example was given in standard form: 2.Factor out the a from the first trinomial and the b from the second trinomial. Then create two perfect square trinomials using the technique of completing the square to obtain a(x – h) 2 + b(y – k) 2 = #. If the ellipse is given in general form, convert it to standard form before graphing. we will need to 3.Divide by the number on the RHS to obtain the standard form, Next Slide

5.3 Ellipses 12 Example 4.Write the given equation of the ellipse in standard form: Group the x terms separately from the y terms and move the constant to the RHS. Complete the square for both trinomials. The numbers added in the parenthesis are 1 and 9. We need to add the same “value” to the RHS. The value is 4 and Write each perfect square trinomial as a binomial squared and add the constants on the RHS. Factor out the ‘4” from the first grouping and the ‘9’ from the second group. Leave a space at the end of each set of parentheses to add the appropriate number when completing the square. Finally, divide by 36 on both sides and simplify to obtain the ellipse in standard form. 36 Your Turn Problem #4 Write the given equation of the ellipse in standard form:

5.3 Ellipses 13 x y The vertices (endpoints of the major axis) are (-1,1) and (5,1). (2,1) 1 st write in standard form using completing the square Now we can find the center, vertices and endpoints. (5,1)(-1,1) Next Slide

5.3 Ellipses 14 (-3,-2) (-2,1) (-4,1) x ● ● ● ●● y center: (-3,1) vertices: (-3,4), (-3,-2) endpoints: (-4,1), (-2,1) Answer: (-3,4) ● ● (-3,1) The End B.R Your Turn Problem #5