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# Resubmit- need step by step answer for Ellipse Equation

What is the equation of the ellipse with foci at (0,-3) and (0,3) and the sum of focal radii being 10?

I tried it two different ways and am not sure if either is correct:
First way:
2a=10 so a=5 (sum of focal radii)

Center is (0,0) halfway between foci

c=3

b^2=a^2-c^2 or b^2 = 16

giving me the equation x^2/16 + y^2/25 = 1

Then I tried it like the textbook by using the distance formula:
squareroot of [x² + (y+3)²] + squareroot of [x² + (y-3)²]=10
When I tried to square both sides, I ended up getting it simplified to x²+y²= 41

Obviously totally different answers—not sure which is correct and why the other is incorrect.

### 1 Answer by Expert Tutors

Ray A. | Experienced Math and Physics Prof.Experienced Math and Physics Prof.
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Your second equation is correct but you did not simplify it correctly. As shown below, you will get the same answer but you'll do more work.

squareroot of [x² + (y+3)²] + squareroot of [x² + (y-3)²]=10

This implies: squareroot of [x² + (y+3)²] = 10 - squareroot of [x² + (y-3)²]

Square each side: x² + (y+3)² = (10 - squareroot of [x² + (y-3)²])2

Simplify each side: x² + y2 + 6y + 9 = 100 +  x² + (y-3)² -20*squareroot of [x² + (y-3)²]

Simplify the right further: x² + y2 + 6y + 9 = 100 + x² + y2 -6y + 9 -20*squareroot of [x² + (y-3)²]

Eliminate like terms:  + y2 + 6y + 9 = 100 + + y2 -6y +9 -20*squareroot of [x² + (y-3)²]
Simplify further: 12y - 100 = -20*squareroot of [x² +y2 -6y + 9]
Divide each side by 4: 3y - 25 = -5 *squareroot of [x² +y2 -6y + 9]
Square again: (3y - 25)= 25(x² +y2 -6y + 9)
Simplify: 9y2 - 150y + 625 = 25x2 + 25y2 -150y + 225
Simplify further: 625-225 = 25x2 + 16y2
This yields: 1 = 25x2/400 + 16y2/400
Finally: 1 = x2/16 + y2/25

Thank you.  I still don't understand, as I lost you at the step where you squared the right hand side of the equation:  not sure where the -20 came from and why it is +x^2 instead of negative.  Ugg.  So frustrated!
(a-b)2=a2+b2 -2ab

Now set a to 10 and b to squareroot of [x² + (y-3)²]

Then (10-squareroot of [x² + (y-3)²])2 = 102 + (squareroot of [x² + (y-3)²])2 -2*10*(squareroot of [x² + (y-3)²]) = 100 + x² + (y-3)² -20*squareroot of [x² + (y-3)²]

Thank you for explaining again.  Much appreciated!