Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
4. Ellipses
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Exercise 40 Page 643

For vertical ellipses the foci are (0,± c). For horizontal ellipses the foci are (± c,0). The value of c can be found by solving the equation c^2=a^2-b^2, where a and b are the absolute values of the nonzero coordinate of the vertices and co-vertices, respectively.

(0,± 2sqrt(7))

Practice makes perfect
Let's start by recalling the general equation of horizontal and vertical ellipses. ccc Horizontal Ellipse& & Vertical Ellipse [0.5em] x^2/a^2+y^2/b^2=1 & & x^2/b^2+y^2/a^2=1 For both equations above a and b are positive numbers, with a>b. Moreover, ± a and ± b are the nonzero coordinates of the vertices and co-vertices, respectively. To find the foci of the ellipse whose equation is given, we will rewrite the equation to match one of the above formats.
36x^2+8y^2=288
â–¼
Simplify
36x^2+8y^2/288=1
36x^2/288+8y^2/288=1
x^2/8+8y^2/288=1
x^2/8+y^2/36=1

a = ( sqrt(a) )^2

x^2/(sqrt(8))^2+y^2/(sqrt(36))^2=1
x^2/(sqrt(4* 2))^2+y^2/(sqrt(36))^2=1
x^2/(sqrt(4)* sqrt(2))^2+y^2/(sqrt(36))^2=1
x^2/( 2sqrt(2))^2+y^2/6^2=1
Since 2sqrt(2) ≈ 2.8 and therefore 6>2sqrt(2), the given equation represents a vertical ellipse.
Equation 36x^2+8y^2=288
⇕
x^2/( 2sqrt(2))^2+y^2/6^2=1
Type of Ellipse Vertical
Vertices (0,± 6)
Co-vertices (± 2sqrt(2),0)
Foci (0,± c)
The value of c, which is the absolute value of the nonzero coordinate of the foci, can be expressed in terms of a and b. c^2=a^2-b^2 Let's substitute 6 for a and 2sqrt(2) for b in the above equation so that we can solve for c.
c^2=a^2-b^2
c^2= 6^2-( 2sqrt(2))^2
â–¼
Solve for c
c^2=6^2-2^2(sqrt(2))^2
c^2=36-4(sqrt(2))^2
c^2=36-4(2)
c^2=36-8
c^2=28
c=sqrt(28)
c=sqrt(4 * 7)
c=sqrt(4) * sqrt(7)
c=2sqrt(7)
Note that in the last step we took the principal root, because c is a positive number. The foci of the given ellipse are (0,2sqrt(7)) and (0,- 2sqrt(7)).