HopelessMathStudent
What are the foci of the hyperbola given by the equation 16x2 - 25y2 - 64x - 50y - 361 = 0?
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helder_edwin
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complete the squares
HopelessMathStudent
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(-1 ± , 2)
(2 ± , -1)
(-1, 2 ± )
(2, -1 ± )
HopelessMathStudent
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Its one of these lol Im just not sure which
jishan
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|dw:1355058848205:dw|
helder_edwin
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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
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got this?
HopelessMathStudent
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Kinda but what are the blank spaces?
helder_edwin
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there's a "y" missing next to the 50.
HopelessMathStudent
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Pretty sure its A or B right?
helder_edwin
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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
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got this?
helder_edwin
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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
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got this?
helder_edwin
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now the center of the hyperbola is
\[ \large (\color{red}{2},\color{red}{-1}) \]
helder_edwin
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is it a horizontal or vertical hyperbola?
how can u tell?
HopelessMathStudent
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the format of the equation.
helder_edwin
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no.
helder_edwin
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look at the denominators which is larger? under x? or under y?
helder_edwin
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yes. this tells u the hyperbola is horizontal
HopelessMathStudent
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I said it was horizontal. Is it B?
helder_edwin
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what is the length of the semi-mayor axis?
helder_edwin
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it HAS to. do u know why?
HopelessMathStudent
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because the formula is formatted (h±a,k) for horizontal and that fits? lol
helder_edwin
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yes. very clever.
u were looking for the foci not the vertices, though.
helder_edwin
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did u understand?
HopelessMathStudent
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do it is B? :) and yeah pretty much lol
helder_edwin
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i hope so.