Question

Find the equation of the ellipse Center at (2,0), Focus (2,6), Eccentricity 2/3. Can someone help me please? Thanks. :)

Answer 1

oops, I blew it?

Answer 2

eccentricity is one that never came up i class ... i know it has something to do with a and b tho

Answer 3

c/a is what the interweb tells me

Answer 4

e = sqrt(1- a^2/b^2)

Am wondering if I have to flip them because ellipse is rotated.

Might have to go back to basics, bring back to to (0,0) calculate then retranslate rotate...yuk!

Answer 5

c is your focal length and a is your vertex length; and b^2 = a^2-c^2 to find the short vertex

Answer 6

rotate? no; just

x^2 y^2
--- + --- = 1
a^2 b^2

a and b can change depending on the major axis

Answer 7

e = sqrt(1- a^2/b^2)

Am wondering if I have to flip them because ellipse is rotated.

Might have to go back to basics, bring back to to (0,0) calculate then retranslate rotate...yuk!

Answer 8

The llipse she is describing is rotated on end, so you have to swap a, b.

Answer 9

since this is elongated along the y axis; a goes under y

Answer 10

I answered it once, guess I got it wrong though.

Answer 11

e = sqrt(1- a^2/b^2)

Am wondering if I have to flip them because ellipse is rotated.

Might have to go back to basics, bring back to to (0,0) calculate then retranslate rotate...yuk!

Answer 12

The basic equation isn't a problem, it's actually calculating the a and b with eccentricity and focus info.

Answer 13

Center (2,0),
Focus (2,6), focal length = 6, which simplifies

Eccentricity 2(3)/3(3) = 6/9

(x-2)^2 (y)^2
------ + ----- = 1
b^2 81

b^2 = 9^2-6^2
= 81-36
= 45 right?