Question

What are the vertices of the hyperbola given by the equation 16x2 – y2 + 96x + 10y + 103 = 0?

(-2, 5) and (-4, 5)

(1, 5) and (-7, 5)

(-3, 6) and (-3, 4)

(-3, 9) and (-3, 1)

Answer 1

Someone Please Help!!!!

Answer 2

Try to complete the square
16x^2 + 96x + 144 = 16(x + 3)^2
y^2 + 10y + __ = (-y + 5)^2

Answer 3

confused... I dont know how.

Answer 4

on the wrong track, sorry

Answer 5

can you help me? Do you know how to answer this?

Answer 6

I think what you need is a form like
\[\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\]

Answer 7

That's why I was trying to complete the square (and did, for x).
I ran into a little trouble with y.

Answer 8

What did you get for x?

Answer 9

16x^2 + 96x + 144 = 16(x + 3)^2

Answer 10

Can you plug x back into the equation and find y?

Answer 11

No

Answer 12

Another possibility is to try the proposed solutions.
If you plug in (-2,5) do you get 0?

Answer 13

I will try all of them, and let you know what works!!!!

Answer 14

Clearly (1, 5) and (-7, 5) doesn't work b/c the first term is definitely not a zero.

Answer 15

It is the first one, Thank you!!!

Answer 16

yw