Question

What are the vertices of the hyperbola given by the equation y^2 – 9x^2 + 22y – 72x – 32 = 0?
(-3, -11) and (-5, -11)

(-1, -11) and (-7, -11)

(-4, -8) and (-4, -14)

(-4, -10) and (-4, -12)

Answer 1

complete the squares first!

Answer 2

y^2+81x+22y-72y-32=0?

Answer 3

let's see. u have
\[ \large y^2-9x^2+22y-72x-32=0 \]
\[ \large (y^2+22y+\qquad)-9(x^2+8x+\qquad)=32 \]

Answer 4

what numbers would u put in the blanks???

Answer 5

@Bunnijjane ???

Answer 6

the second blank is 5.65 i think. I was trying to figure out the first.

Answer 7

no

Answer 8

for the first it would be
\[ \large \left(\frac{22}{2}\right)^2=? \]

Answer 9

121?

Answer 10

yes. now do the same for the second part!!

Answer 11

16?

Answer 12

great!!

Answer 13

now we have
\[ \large (y^2+22y+11^2)-9(x^2+8x+4^2)=32+\qquad-\qquad. \]

Answer 14

u have to put two numbers in the blanks. which are they?

Answer 15

the ones u just added!!

Answer 16

wow I'm a moron -_-. Sorry! so now the equation is (y^2+22y+11)-)x^2+8x+4^2)=32+121-12?

Answer 17

the first is correct.

Answer 18

the second one would be
\[ \large -9(16)=-144 \]

Answer 19

do u get this??

Answer 20

so the equation would be
\[ \large (y+11)^2-9(x+4)^4=32+121-144=9 \]
dividing by 9 we have
\[ \large \frac{(y+11)^2}{9}-\frac{(x+4)^2}{1}=1 \]

Answer 21

The answer is c.(-4, -8) and (-4, -14)