anonymous
  • anonymous
What are the foci of the hyperbola given by the equation 16x2 - 25y2 - 64x - 50y - 361 = 0?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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helder_edwin
  • helder_edwin
complete the squares
anonymous
  • anonymous
How?
anonymous
  • anonymous
(-1 ± , 2) (2 ± , -1) (-1, 2 ± ) (2, -1 ± )

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anonymous
  • anonymous
Its one of these lol Im just not sure which
anonymous
  • anonymous
|dw:1355058848205:dw|
helder_edwin
  • helder_edwin
first factor the coeficientes of the squared terms (i re-arranged the terms) \[ \large 16x^2-64x-25y^2-50=361 \] \[ \large 16(x^2-4x+\qquad)-25(y^2+2y+\qquad)=361 \]
helder_edwin
  • helder_edwin
got this?
anonymous
  • anonymous
Kinda but what are the blank spaces?
helder_edwin
  • helder_edwin
there's a "y" missing next to the 50.
anonymous
  • anonymous
oh okay
anonymous
  • anonymous
Pretty sure its A or B right?
helder_edwin
  • helder_edwin
now in each parenthesis there's a blank. u have to write there coeficient of the linear term (no power) halved and then squared \[ \large \color{green}{16}(x^2-4x+\color{red}{4})\color{orange}{-25}(y^2+2y+\color{red}{1})=361 +\color{green}{16}\cdot\color{red}{4}\color{orange}{-25}\cdot\color{red}{1} \]
helder_edwin
  • helder_edwin
got this?
helder_edwin
  • helder_edwin
http://en.wikipedia.org/wiki/Hyperbola
anonymous
  • anonymous
mhm :)
helder_edwin
  • helder_edwin
now \[ \large 16(x-\color{red}{2})^2-25(y\color{red}{+1})^2=400 \] \[ \large \frac{16(x-\color{red}{2})^2}{400}-\frac{25(y-(\color{red}{-1}))^2}{400}=1 \] \[ \large \frac{(x-\color{red}{2})^2}{25}-\frac{(y-(\color{red}{-1}))^2}{16}=1 \]
helder_edwin
  • helder_edwin
got this?
anonymous
  • anonymous
I see :)
helder_edwin
  • helder_edwin
now the center of the hyperbola is \[ \large (\color{red}{2},\color{red}{-1}) \]
helder_edwin
  • helder_edwin
is it a horizontal or vertical hyperbola? how can u tell?
anonymous
  • anonymous
horizontal
anonymous
  • anonymous
the format of the equation.
helder_edwin
  • helder_edwin
no.
helder_edwin
  • helder_edwin
look at the denominators which is larger? under x? or under y?
anonymous
  • anonymous
x.....
helder_edwin
  • helder_edwin
yes. this tells u the hyperbola is horizontal
anonymous
  • anonymous
I said it was horizontal. Is it B?
helder_edwin
  • helder_edwin
what is the length of the semi-mayor axis?
helder_edwin
  • helder_edwin
it HAS to. do u know why?
anonymous
  • anonymous
because the formula is formatted (h±a,k) for horizontal and that fits? lol
helder_edwin
  • helder_edwin
yes. very clever. u were looking for the foci not the vertices, though.
helder_edwin
  • helder_edwin
did u understand?
anonymous
  • anonymous
do it is B? :) and yeah pretty much lol
anonymous
  • anonymous
so*
helder_edwin
  • helder_edwin
i hope so.

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