## anonymous 5 years ago The vertices of the hyperbola 25^(2)-9x^(2)-200+36+139=0 are a) (-1,4) and (5,4) b) (2,9) and (2,-1) c) (2,7) and (2,-1) d) (2,9) and (2,-1)

1. anonymous

25y^(2)-9x^(2)-200y+36x+139=0***

2. anonymous

hi chizzle, make them complete squares .This will give you (5y+20)^2 - (3x -6)^2 -100 +36 +139 = 0. This form should now look familiar.(standard hyperbola form).

3. anonymous

p.s. there might be a calculation error so check it once but I'm sure this is the general idea. you then can find out how much the graph has been shifted by looking a the modifiers for x and y .i.e x-> x-a; y->y-b

4. anonymous

hi chizzle, make them complete squares .This will give you (5y+20)^2 - (3x -6)^2 -100 +36 +139 = 0. This form should now look familiar.(standard hyperbola form).

5. anonymous

What I got before was $2, 4\pm \sqrt{171/5}$ but that is not one of the answers. I need in the form of coordinates: (-1,4) and (5,4) (2,9) and (2,-1) (2,7) and (2,-1)

6. anonymous

hint: centre = (2,-5) ;a = 3 ;b = 5