equation of an Ellipse?

Question

What is the equation of the ellipse with foci at (0,-4) and (0,4) and the sum of it's focal radii being 10?

Answer 1

GOT IT

these are the series of steps, i've verified the answer and it works!, i hope you are pleased with this...

use length of a line formula

sqrt[x^2 + (y+4)^2] + sqrt[x^2 + (y-4)^2] = 10

move to other side

sqrt[x^2 + (y+4)^2] = 10 - sqrt[x^2 + (y-4)^2]

square both sides and expand accordingly

x^2 + y^2 + 8y + 16 = 100 -20sqrt[x^2 + (y-4)^2] + x^2 + y^2 -8y + 16

cancel like terms

8y = 100 -20sqrt[x^2 + (y-4)^2] - 8y

move 8y term

16y = 100 -20sqrt[x^2 + (y-4)^2]

common factor 20

16y = 20(5 - sqrt[x^2 + (y-4)^2])

divide by 20 each side

(4/5)y = 5 - sqrt[x^2 + (y-4)^2]

multiply each term by 5 for cleaning the fraction on the left side

4y = 25 - 5sqrt[x^2 + (y-4)^2]

move 25 to left side

4y - 25 = - 5sqrt[x^2 + (y-4)^2]

square both sides once again

16y^2 -200y + 625 = 25(x^2 + y^2 -8y +16)
16y^2 -200y + 625 = 25x^2 + 25y^2 -200y + 400

cancel terms and collect like terms

-9y^2 - 25x^2 = -225

remove minus sign for ease of use

9y^2 + 25x^2 = 225

divide through everything by 225 to get useful equation

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

this is the useful equation

therefore this ellipse has center (0, 0)

a = 5 (major axis)
b = 3 (minor axis)
c = 4 (foci)

a^2 = b^2 + c^2 is satisfied...


i know user helmut got it right, but if this were to be a nastier question with a center of the ellipse being somewhere in the decimals and fractions, his method wouldn't work....this is the actual correct way to do it. Other than that, everything looks good.

Answer 2

c^2 = b^2 - a^2
c = 4, b = 5, a = 3
x^2/9 + y^2/25 = 1