Question

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)

Answer 1

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

Answer 2

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).

Answer 3

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

Answer 4

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).

Answer 5

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)

Answer 6

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