Question

Find the center and vertices of the hyperbola.
11x^2 - 25y^2 - 110x + 150y - 225 = 0

Answer 1

lordamercy
you have to first write is in standard form as
\[\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\] so you know how to do that?

Answer 2

no

Answer 3

ok well lets get busy, because frankly you can't really eyeball it, you have to complete the square twice, once for the \(x\) terms and once for the \(y\) terms

Answer 4

we have \[11x^2 - 25y^2 - 110x + 150y - 225 = 0\] or
\[11x^2 - 25y^2 - 110x + 150y = 225
\]

Answer 5

lets put the \(x\) terms together to get
\[11x^2-110x-25y^2+150y=225\] and then factor out the common factors for the \(x\) terms and the \(y\) terms and get
\[11(x^2-10x)-25(y^2-6y)=225\]

Answer 6

how we doing so far?

Answer 7

im keeping up so far

Answer 8

ok so now to "complete the square"
for the term \(11(x^2-10x)\) we think as follows
half of 10 is 5, 5 squared is 25 so we want to write it as
\[11(x^2-10x+25)=11(x-5)^2\]
since we added \(11\times 25=275\) to the left, we have to add that same number to the right

Answer 9

now we have
\[11(x-5)^2-25(y^2-6y)=225+275=500\]

Answer 10

we have to repeat the process for the \(y\) terms
again we think
half of 3 is 3, and 3 squared is 9 so we write
\[-25(y^2-6y+9)=-25(y-3)^2\] this time we subtracted \(25\times 9=225\) from the left, so we have to subtract it also from the right

Answer 11

now we got
\[11(x-5)^2-25(y-3)^2=275\]

Answer 12

that was the only real hard part

Answer 13

now to make it look like this \[\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\] divide both sides by \(275\) to get
\[\frac{(x-5)^2}{25}-\frac{(y-3)^2}{11}=1\]

Answer 14

NOW we can actually answer the question
the center is displayed for you know since the center of \[\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\] is \((h,k)\)

Answer 15

let me know if i lost you
otherwise we should essentially be done

Answer 16

thank you so much!

Answer 17

yw
now that we got the center at \((5,3)\) do you know how to find the vertices?

Answer 18

yeah, i should be good. thanks again!

Answer 19

yw

Answer 20

actually i tried to get the vertices but kept getting messed up. could you explain?