SOLUTION: Graph: (x+3)^2+(y+1)^2=4 (y+2)^2=4(4)(1)(x-1) (x+2)^2/9-(y-3)^2/4=1 (x-2)^2/4+(y+3)^2/9=1 Thanks soo much

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Graph: (x+3)^2+(y+1)^2=4 (y+2)^2=4(4)(1)(x-1) (x+2)^2/9-(y-3)^2/4=1 (x-2)^2/4+(y+3)^2/9=1 Thanks soo much      Log On


   

Question 611284: Graph:

(x+3)^2+(y+1)^2=4


(y+2)^2=4(4)(1)(x-1)

(x+2)^2/9-(y-3)^2/4=1

(x-2)^2/4+(y+3)^2/9=1

Thanks soo much
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

(x+3)^2+(y+1)^2=4   ||Circle with C(-3,-1) and radius of 2







 (y+2)^2=16px-1)   ||The standard form is %28x+-h%29%5E2+=+4p%28y+-k%29, where  the focus is (h,k + p)

parabola opening to the right, C(1,-2) and focus(5,-2) 4p=16, p = 4





(x+2)^2/9-(y-3)^2/4=1 

Standard Form of an Equation of an Hyperbola is  %28x-h%29%5E2%2Fa%5E2+-+%28y-k%29%5E2%2Fb%5E2+=+1 opening right or left

 where Pt(h,k) is a center  with vertices 'a' units right and left of center.

 Asymptotes passing thru the center with slope = ± b/a 

foci being ± sqrt(a^2 + b^2) from center  along axis of symmetry y = b



%28x-2%29%5E2%2F2%5E2%2B%28y%2B3%29%5E2%2F3%5E2=1     ||Ellipse with C(2,-3) V(2,0) and V(2,-6)