1003024101

SubArea: 
Conics
Level: 
B
Project ID: 
1003024101
Source Problem: 
2010006306
Accepted: 
1
Clonable: 
1
Easy: 
0
The equation of a hyperbola with the center \( S=[-1;3] \), the focus \( F=[4;3] \), and the vertex \( A=[2;3] \) is given by:
\( \frac{(x+1)^2}{9}-\frac{(y-3)^2}{16} =1 \)
\( \frac{(x-1)^2}{9}-\frac{(y+3)^2}{16} =1 \)
\( \frac{(x+1)^2}{9}+\frac{(y-3)^2}{16} =1 \)
\( \frac{(x-1)^2}{16}-\frac{(y+3)^2}{9} =1 \)
\( \frac{(y-3)^2}{16}-\frac{(x+1)^2}{9} =1 \)