2010006306

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