HELP! Algebra 2 problems (Not my homework; school starts in a week for me:September 6)?

Question

Please help me. And please show/tell me how you got the answer


1) Find the center of a circle with the equation x^2 + y^2 + 2x + 4y - 9 = 0

2) Find the foci of a hyperbola with the equation 9y2 - 72y - 16x2 - 64x - 64 = 0.
a. (3, 4) and (-7, 4)
b. (-3, -4) and (7, -4)
c. (2, 1) and (2, -9)
d. (-2, -1) and (-2, 9)


3) Which equation is shown by the graph in Figure 4?

Figure 4 opens up and down, with the two vertex at (0,4) and (0,-4)

a. (y^2) / 16 - (x^2) / 9 = 1

b. (y^2) / 9 - (x^2) / 16 = 1

I only need the answer to #2 now, thanks guys and girls

Answer 1

(1) (x^2+2x+1)+(y^2+4y+4)=16
(x^2+1)^2+(y^2+2)^2=16
so i think the center is (-1, -2)
(i'm not sure i just learned it, i forgot through the summer.......)
(2)u plug all the answers in and then figure out which one fits. there's surelly a easier way, but since i can't solve in a easy way so............
(3)i don understand ur graph, send me E-mail if u want, i can help u.

Answer 2

Uhhh.... ok, since #3's easy, its A cause the distance between the two vertices is 8 (i got it from 4 minus (-4) which equals 4+4=8). Since the the general equation of the hyperbola has the denominator of the y^2 term being the square of the distance between the center of the hyperbola (which we know is (0,0)) and the vertex (which is half of 8, so its 4), the denominator must be 4^2 which is 16.

Basically, count from the center of a up/down hyperbola to any vertex. Take this distance, square it, and you get the number under the y^2.

the others I could do but im really lazy. so sorry.

Answer 3

Don't you own a Graphing Calculator Yet? They are not that expensive and a must have item. Just plug in the numbers for the correct answer. It's a very simple equation.

Answer 4

Ok so #2.
You have to plug in the x and y values. For a... x=3 and y=4.... then x=-7 y=4.

Plug em up and see what ya get!

Answer 5

sure...