SOLUTION: find center, foci, vertices, and asymptotes of the hyperbola.
16x^2-9y^2-96y+144=0
somehow i have gotten it down to:
(16(x-3)^2)/0-(9y^2)/0=1
Algebra.Com
Question 722706: find center, foci, vertices, and asymptotes of the hyperbola.
16x^2-9y^2-96y+144=0
somehow i have gotten it down to:
(16(x-3)^2)/0-(9y^2)/0=1
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
I cannot emphasize this strongly enough: You cannot, under any circumstances, divide by zero!
Now, assuming that the 
in your original is a typo and you really meant 
, then everything else you did was correct. You should have obtained:
^2\ -\ 9y^2\ =\ 0)
This is a degenerate hyperbola, in other words two intersecting straight lines. The "center" is the point of intersection. There are no vertices or foci, and the lines themselves would be the asymptotes if the RHS of your equation were other than zero.
^2\ =\ 9y^2)
Take the square root of both sides:
\ =\ \pm3y)
Which are the two straight lines:
Which intersect at )
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
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