An ellipse has this Cartesian equation:

Question

((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

Answer 1

I 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.