easy question.....?

Question

Give me its solution how I can do it....

Find the center, vertices, and foci for the ellipse: x^2/36 + Y^2/81 = 1

Answer 1

I hate mathematics... Brr... I can't help you, sorry

Answer 2

General formula for an ellipse is:
(x-h)^2/(a^2) + (y-k)^2/(b^2) = 1 where b <=a.

Major axis has equation y=k, minor axis has equation x=h.

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

Vertices are the two endpoints of the major axis, each located a distance "a" from the center. Foci are the endpoints of the minor axis, each located a distance "a * e" from the center.

Therefore, from your equation:
(x^2) = (x-h)^2 ... h=0
(y^2) = (y-k)^2 ... k=0
36 = (a^2) ... a=6, a=-6
81 = (b^2) ... b=9, b=-9
e = sqrt (a^2 - b^2)/a = sqrt(36-81)/6 = 1.118

"b" must be less than or equal to "a", therefore b = -9.

Center is at (h,k) = (0,0).
Vertices are located at (a,0) and (-a,0) ... which means (6,0) and (-6,0).
Foci are located at (0,a*e) and (0,-a*e) ... which means (0,6.708) and (0,-6.708).

Answer 3

Sorry but that is not a easy question. Plus, i am not that good at math but the basic stuff!

Answer 4

Easy huh? I would think if that were the case, you would have solved it. Sorry, math is not my forte... it's not an easy question for me.

Answer 5

5.7?; ==b34e?+90=^x7=(+9/65-345[8+6^]*h5t5-7?/rm

/,.>p8=
23454[p]=p+(89_)8*> this is just an aproximation

Answer 6

read all about it here --> http://en.wikipedia.org/wiki/Ellipse