Question

Pre-Cal! hyperbola and ellipses.

x^2+9y^2-4x-72y+139=0

find the: center, vertices, foci, main axis.
How would I go about solving this? Can anyone help and explain how what to do? Im having the most trouble understanding how to get the foci.

Answer 1

first you would go about completing the square: 1. (x^2 - 4x + _) + (9y^2- 72y + _) = -139

by adding to both sides the number that will make these perfect squares
2. (x^2 -4x+4) + 9(y^2-8y +16) = -139 +4 + 9(16)

then you would factor so it would look like this: 3. (x-2)^2 +9(y-4)^2 = 9

side note:You can tell that this is going to be an ellipse because its addition, if it was a hyperbole it would look like this :[(x-2)^2 -9(y-4)^2 = 9].

Anyhow, now you"re gunna divide the equation by 9 so that the left hand side of the equation is 1, which would give you the standard form of the equation.: [(x-2)^2]/9 + [(y-4)^2]/1 =1

to help this make more sense, here is a general outline of the standard equation for an ellipse:
[(x-h)^2]/a^2 + [(y-k)^2]/b^2 =1 (h and k are your center)

Now you get your values of a and b, which will give you your vertices. for an ellipse, a is always the square root of the largest dividing number. So a^2 =9, and b^2=1. which means a is 3, and b is 1.

your center is going to be (2,4)

you can tell that this equation is going to be along the x axis (horizontal) becaus a^2 is dividing x. if it was underneath y than it would vertical.

if your center was at (0,0), your vertices would simply be (a,0) and (-a,0) ---> ( 3,0) and (-3,0)

but because your center is (2,4) you have to "add" in a way, the center to your vertices:
(3,0) + (2,4) = (5,4) and (-3,0) + (2,4) = (-1,4).

now you are also going to have to do the whole "adding" thing for your foci, but first you have to find c. for an ellipse, you can find c with the equation c^2 =a^2-b^2.

so, c^2=9-1 ---> c = +/- rad 8 (I dont have a square root key so "rad" will have to do)

your foci is always on the some axis as a is, so if your center was (0,0), the foci would look something like this (rad8,0) and (-rad8,0)

after you add the center to it, your foci would become: (2+rad8,4) and (2-rad8,4)

tell me if you have any questions

Answer 2

Wow! Thanks a lot! You really helped so much! I'm still trying to get the hang of these problems! So as for the main axis would I just do "2a"? so, 2(9)=18?