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August 30, 2014

August 30, 2014

Posted by **Grace** on Friday, January 20, 2012 at 12:44am.

((x-2)/5)^2 + ((y-4)/3)^2 = 1

Find the focal radius

Find the eccentricity

Write the parametric equations for the ellipse

Transform into the form, Ax^2 + By^2 + Cx + Dy + E = 0

- Precalc -
**Reiny**, Friday, January 20, 2012 at 9:51amI don't know which author you use, but focal radius can be defined in more than one way

http://www.mathwords.com/f/focal_radius.htm

for yours,

a = 5, b = 3

your ellipse will be horizontal, and

b^2 + c^2 = a^2

9+c^2 = 25

c^2 = 16

c = 4

eccentricity = c/a = 4/5 = .8

parametric equations:

x-2 = 5coss t ---> x = 5cost + 2

y-4 = 3sin t ----> y = 3sint + 4

original:

(x-2)^2 /25 + (y-4)^2 / 9 = 1

multiply by 225

9(x-2)^2 + 25(y-4)^2 = 225

9(x^2 - 4x + 4) + 25(y^2 - 8y + 16) = 225

I am sure you can finish it.

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