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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))
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.
.LHS /288.=.RHS /288.
Write as a sum of fractions
a/b=.a /36./.b /36.
a/b=.a /8./.b /8.
a = ( sqrt(a) )^2
Split into factors
sqrt(a* b)=sqrt(a)*sqrt(b)
Calculate root
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.
b= 2sqrt(2), a= 6
(a b)^m=a^m b^m
Calculate power
( sqrt(a) )^2 = a
Multiply
Subtract term
sqrt(LHS)=sqrt(RHS)
Rewrite 28 as 4 * 7
sqrt(a* b)=sqrt(a)*sqrt(b)
Calculate root
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)).