Week 10, Day 2

Today we learned about another type of conic: ellipses, a set of all points (x,y) the sum of whose distances from two distinct fixed points (foci) is constant. Its components are illustrated below:

The line through the foci intersects the ellipse at two points (vertices). The chord joining the vertices is the major axis, and its midpoint is the center of the ellipse. The chord perpendicular to the major axis at the center is the minor axis of the ellipse. To derive the standard form of the equation of an ellipse, consider the figure shown below with these points: center (h,k); vertices (h + a, k); foci, (h +/- c, k). The sum of the distances from any point on the ellipse to the two foci is constant. Using a vertex point, this constant sum is (a + c) + (a - c) = 2a (length of major axis) or simply the length of the major axis. Now, if you let (x,y) be any point on the ellipse, the sum of the distances between (x,y) and the two foci must also be 2a. That is, [(x - (h - c))^2 + (y - k)^2]^1/2 + [(x - (h + c))^2 + (y - k)^2]^1/2 = 2a. Finally, using the figure below and b^2 = a^2 - c^2, you obtain the following equation of the ellipse. b^2(x - h)^2 + a^2(y - k)^2 = a^2b^2; (x - h)^2/a^2 + (y - k)^2/b^2 = 1. 

Standard Equations of an Ellipse:

Vertical and Horizontal Orientations of an Ellipse:

Eccentricity:

To measure the ovalness of an ellipse, you can use the concept of eccentricity. 

The eccentricity (e) of an ellipse is given by the ratio: e = c / a. 

Example 1: 

x^2 + 4y^2 + 6x - 8y + 9 = 0

Find the standard form, center, foci, vertex, and eccentricity of this equation. 

x^2 + 6x + 4y^2 - 8y = -9

x^2 + 6x + 9 + 4(y^2 - 2y) = -9 + 9

x^2 + 6x + 9 + 4(y^2 - 2y + 1) = -9 + 9 + 4

(x + 3)^2 + 4(y - 1)^2 = 4

(x + 3)^2 / 4 + (y - 1)^2 / 1 = 1

Center = (h, k)

Center = (-3, 1)

b = 1, a = 2

c^2 = a^2 - b^2

c^2 = 4 - 1

c^2 = 3

c = (3)^1/2

Foci = (-3 + (3)^1/2, 1) (-3 - (3)^1/2, 1)

Vertex = (-1, 1) (-5, 1) 

Eccentricity = e = c / a = (3)^1/2 / 2