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What are the foci of the hyperbola given by the equation 16x2 - 25y2 - 64x - 50y - 361 = 0?

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complete the squares
(-1 ± , 2) (2 ± , -1) (-1, 2 ± ) (2, -1 ± )

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Other answers:

Its one of these lol Im just not sure which
first factor the coeficientes of the squared terms (i re-arranged the terms) \[ \large 16x^2-64x-25y^2-50=361 \] \[ \large 16(x^2-4x+\qquad)-25(y^2+2y+\qquad)=361 \]
got this?
Kinda but what are the blank spaces?
there's a "y" missing next to the 50.
oh okay
Pretty sure its A or B right?
now in each parenthesis there's a blank. u have to write there coeficient of the linear term (no power) halved and then squared \[ \large \color{green}{16}(x^2-4x+\color{red}{4})\color{orange}{-25}(y^2+2y+\color{red}{1})=361 +\color{green}{16}\cdot\color{red}{4}\color{orange}{-25}\cdot\color{red}{1} \]
got this?
mhm :)
now \[ \large 16(x-\color{red}{2})^2-25(y\color{red}{+1})^2=400 \] \[ \large \frac{16(x-\color{red}{2})^2}{400}-\frac{25(y-(\color{red}{-1}))^2}{400}=1 \] \[ \large \frac{(x-\color{red}{2})^2}{25}-\frac{(y-(\color{red}{-1}))^2}{16}=1 \]
got this?
I see :)
now the center of the hyperbola is \[ \large (\color{red}{2},\color{red}{-1}) \]
is it a horizontal or vertical hyperbola? how can u tell?
the format of the equation.
look at the denominators which is larger? under x? or under y?
yes. this tells u the hyperbola is horizontal
I said it was horizontal. Is it B?
what is the length of the semi-mayor axis?
it HAS to. do u know why?
because the formula is formatted (h±a,k) for horizontal and that fits? lol
yes. very clever. u were looking for the foci not the vertices, though.
did u understand?
do it is B? :) and yeah pretty much lol
i hope so.

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