The Ellipse x y r - The Burns Home Page

The EllipseAppolonius of Perga (225 B.C.) discovered that by intersecting a right circular cone all theway through with a plane slanted but is not perpendicular to the axis, the intersectionprovides a resulting curve (conic section) in the shape of a ellipse.Ellipses are commonly noted as the actual paths the planets take when orbiting the sun.2 2 2Its formula is derived from the equation of a circle, x y r , where one of thecoefficients of2xor2yis larger than the other.The Ellipse in Terms of a Locus of Points Version 1An Ellipse is the set of all points such that the sum of the distances from two fixedpoints, called foci, is constant.F 1F 2The two distances from a point on the ellipse to its foci are the focal radii.

The easiest and most available method for drawing an ellipse is with two nails, a pencil,and a piece of string. We begin by trying the string at the two ends to make a loop. Nextwe put the push nails into a wall or a desk (the nails are the foci) and lay the stringaround the nails. Lastly we put the pencil inside the string and pull it tight around thenails. The trace that we produce will be an ellipse (visit first Web Link for animation).The Ellipse in Terms of a Locus of Points Version 2The traditional way to define an ellipse, as a conic section. It is the set of points whosedistance from a point (the focus) and a line (the directrix) is in a constant proportion).

The Anatomy of an EllipseBbV'F'acOFVPB'‣ The point O, the midpoint F’F, is the centre of the ellipse. Each of the foci is cunits from the centre.‣ The points V’ and V are the vertices of the ellipse. The vertices are where theaxis joining the foci (Focal Axis) intercepts the ellipse.‣ The line segment V’V is the major axis. The major axis lies on top of the FocalAxis and is an axis of symmetry. Since each vertex is a units from the centre,therefore the length of the major axis is 2a.‣ The points B’B are the endpoints of the minor axis. The minor axis is the linesegment through the centre of the ellipse perpendicular to the major axis. Thelength of the minor axis is 2b. Note: the minor axis is always shorter than themajor axis.‣ The sum of the distance from the two foci to any point on the ellipse is 2a2 2 2‣ The distances a, b, and c are related by the equation a b c .Some ellipses are more elongated that other ellipses. The closer the foci are to the centrethe more the ellipse becomes circular. We call the ratio of c athe eccentricity (e).The Eccentricity‣ Becausece and 0 c a , then 0 e 1.a‣ The closer e is to 0, the closer the foci are to the centre and the more circularthe ellipse.‣ The closer e is to 1, the closer the foci are to the vertices and the moreelongated the ellipse.

Deriving an horizontal Ellipse Equation with Centre at (0, 0) and foci Let P(x,y) represent any point on the ellipse, and let F 1 and F 2 represent the two foci ofthe ellipse. Then PF 1 + PF 2 =2aNow rewriting PF 1 + PF 2 =2a in terms of distances2 2 2 2 x c y 0 x c y 0 2a2 2 2 2x c y 0 2 a x c y 02 2 2 4 4 x c y a a x c y x c y2 2 2 222x 2 cx c y 4 a 4 a x c y x 2cx c y2 2 2 2 2 2 2 24cx 4 a 4a x c y2 216c x 32ca x 16a 16a x 2cx c y2 2 2 4 2 2 2 22 2 2 4 2 2 216c x 32ca x 16a 16a x 32ca x 16a c 16a y16c x 16a x 16a y 16a c 16a2 2 2 2 2 2 2 2 4c x a x a y a c a2 2 2 2 2 2 2 2 42 2 2 2 2 2 2 2c a x a y a c a2 2 2 2 2 2b x a y a b b x a y a bxa2 2 2 2 2 2yb2 22 21a>b2 2 2 2c a b2 2 2c,0If we place a radical oneither side it is easier tosquare and remove one ofthe radical expressions.Again get the radical by itselfso squaring will eliminateradical sign.Similarly the Vertical Ellipse equation isxbya2 212 2 with foci at 0, c , and a>bThe standard form of the equations for the ellipse whose centre is (h,k) isx h 2 y k2 , for a horizontal major axis, where a>b, and foci hc , k ab2 21x h 2 y k2 , for a vertical major axis, where a>b, and foci h , k cba2 21

ExampleDescribe the key properties ofSolution:2 2x y25 161Centre and Orientation of the EllipseThe ellipse is in the formcentre (0, 0).xayb2 2 12 2 with a>b, then this is an Horizontal ellipse withLengths of Major and Minor AxisSince a 2 25and b 2 16 , we have a 5and b 4 . Therefore the length of the Majoraxis isa , and the Minor Axis is b 2 2 5 102 2 4 8x-interceptsTo find the x-intercepts (vertices, since origin is (0, 0)), set y=0 and solve for x.x 2 so x 5251Hence the x-intercepts are 5,0and 5,0y-interceptsTo find the y-intercepts set x=0 and solve for y.y 2 , so y 4161Hence the y-intercepts are and 0,40, 4HorizontalEllipseLocation of FociSince c 2 a 2 b2 25 16 9 , c=3. The foci are on the x-axis, 3 units on either side ofthe origin at 3,0 and 3,0

ExampleFind the equation of the ellipse with centre (0,0), foci (5,0) and (-5,0) and y-intercepts 6and –6.Solution:The foci are on the x-axis. Therefore, the major axis is horizontal.The equationxayb2 2 12 2 , where a>b, and b=6 and c=5. Find aa b c2 2 26 52 261a 61y-interceptsare 6 and -6Foci at 5and -5Therefore2 2x y 161 36ExampleFind the centre and the length of the major and the minor axes for the ellipse with the2 2following equation: x ySolution:4 2 16 1 64Change into standard form by dividing by 612 2x y4 2 16 1 64 64 64 642 2x2 y116 4The centre of the ellipse is (2,-1), and a=4 and b=2The length of the major axis is 2a=2(4)=8, and the length of the minor axis is2b=2(2)=4.1ExamplewidthThe shape of an arch is a semi-ellipse, and its span is the major axis of the ellipse. If thespan is 60m and its height is 20m, find the height of the arch at a point 12m from thecentre, to the nearest tenth of a metre.

Solution:With these type of archway question, it is best to make the centre of the road the origin.Then we have half the road of the left and the other half on the right of the centre (0, 0)with the tallest height above the middle of the road.Letting the point (12, y) be on the ellipse, then we need only determine y.Since this is a horizontal ellipse, we havexay with a=30, b=20b2 22 21Thus the equation of the ellipse is2 2x y 1900 400Substitute x=12 in the equation, and solve for y 2 212 y 1900 4002144 y 1900 4002y 1441400 9002y 21400 252y 336y 4 21Since y must be positive, the height of the arch is approximately 18.3m

ExampleDetermine the vertices and foci for the following ellipse2 23x2 y 18Solution:Rewrite equation of ellipse in the formxby , by dividing by 18.a2 212 2Vertical Ellipse2 23x2 y 182 2x y6 91Therefore a 6 and b 3.Since 9>6 we have a vertical ellipse with vertex at (0, 3) and (0, -3)Now for an ellipse we know a 2 b 2 c2c9 6c23c 32The foci are: 0, 3 , 0, 3Example 2 2x 1 y 3Determine the vertices and foci for the following ellipseSolution:This is a horizontal ellipse with centre h , k 1,3Alsoa b c2 2 2100 36cc264c 82Foci are: hc , k 18,3 9,3 , 7,3Vertices are: ha , k 110,3 11,3 , 9,3100 361

ExampleDetermine the equation of the ellipse with major axis 32, centre 1,7 .Solution:1, 2, and one focus atWe know h , k 1, 2and that since the x-value of both the centre and the focus arethe same (1), then this is a vertical ellipse.y value of vertexWe also know that 2a=32, or a=16 and that Nowa b c 2 2 22 2 216 b9b2175h,k cso 7 2 cc 9Only value ofc possibleThe general equation isxb 2 2x 1 y 2, thereforeya2 22 21175 2561ExampleThe vertices of an ellipse with eccentricity 0.8 are located at the standard form of the equation of the ellipse.Solution:2,2and 8,2 . DetermineFirst determine if the ellipse is horizontal or vertical. Since the vertices have the same y-coordinate but different x-coordinates, the vertices lie on a horizontal line, so this is ahorizontal ellipse.The centre (h, k) of the ellipse is on the line that contains the vertices and is midwaybetween them. 2 8 2 2 2 2 Therefore h , k, 3,2Let’s find a: The distance between (3, 2) and (8, 2) is 5 units thus a 5and2a 25Find c: sincece ac0.8 5c 4

Now for b:a b c2 2 2225 b16b29 2 2x h y kPutting it all together :ab2 22 2x3 y225 911