An ellipse has this Cartesian equation:

((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

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.

To find the focal radius of an ellipse, you need to determine the distance between the center of the ellipse and one of its foci. The formula for the focal radius (r) is given by the equation:

r = √(a^2 - b^2), where a is the length of the semi-major axis and b is the length of the semi-minor axis.

In the given Cartesian equation of the ellipse: ((x-2)/5)^2 + ((y-4)/3)^2 = 1, we can determine the values of a and b:
a = 5 (since the semi-major axis is the square root of the denominator of the x term)
b = 3 (since the semi-minor axis is the square root of the denominator of the y term)

Using the formula for the focal radius, we can calculate it as follows:
r = √(a^2 - b^2)
= √(5^2 - 3^2)
= √(25 - 9)
= √16
= 4

Therefore, the focal radius of the ellipse is 4.

To find the eccentricity of an ellipse, you can use the formula:
e = √(1 - (b^2/a^2)), where a is the length of the semi-major axis and b is the length of the semi-minor axis.

Using the given values of a = 5 and b = 3, we can calculate the eccentricity:
e = √(1 - (3^2/5^2))
= √(1 - 9/25)
= √(16/25)
= 4/5

Therefore, the eccentricity of the ellipse is 4/5.

To write the parametric equations for the ellipse, we can use the trigonometric parameterization. The parametric equations are given by:
x = a*cos(t)
y = b*sin(t), where t is the parameter that ranges from 0 to 2π.

Using the values of a = 5 and b = 3, the parametric equations for the ellipse become:
x = 5*cos(t)
y = 3*sin(t)

To transform the ellipse into the form Ax^2 + By^2 + Cx + Dy + E = 0, we need to expand and rearrange the given equation:
((x-2)/5)^2 + ((y-4)/3)^2 = 1
Simplifying this equation, we get:
(x-2)^2/25 + (y-4)^2/9 = 1
Expanding and rearranging, we obtain:
9(x-2)^2 + 25(y-4)^2 - 225 = 0

Hence, the transformed equation in the required form is:
9x^2 - 36x + 225y^2 - 900y + 1864 = 0.