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HopelessMathStudent

  • 2 years ago

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

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  1. helder_edwin
    • 2 years ago
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    complete the squares

  2. HopelessMathStudent
    • 2 years ago
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    How?

  3. HopelessMathStudent
    • 2 years ago
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    (-1 ± , 2) (2 ± , -1) (-1, 2 ± ) (2, -1 ± )

  4. HopelessMathStudent
    • 2 years ago
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    Its one of these lol Im just not sure which

  5. jishan
    • 2 years ago
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    |dw:1355058848205:dw|

  6. helder_edwin
    • 2 years ago
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    first factor the coeficientes of the squared terms (i re-arranged the terms) \[ \large 16x^2-64x-25y^2-50=361 \] \[ \large 16(x^2-4x+\qquad)-25(y^2+2y+\qquad)=361 \]

  7. helder_edwin
    • 2 years ago
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    got this?

  8. HopelessMathStudent
    • 2 years ago
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    Kinda but what are the blank spaces?

  9. helder_edwin
    • 2 years ago
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    there's a "y" missing next to the 50.

  10. HopelessMathStudent
    • 2 years ago
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    oh okay

  11. HopelessMathStudent
    • 2 years ago
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    Pretty sure its A or B right?

  12. helder_edwin
    • 2 years ago
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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 \[ \large \color{green}{16}(x^2-4x+\color{red}{4})\color{orange}{-25}(y^2+2y+\color{red}{1})=361 +\color{green}{16}\cdot\color{red}{4}\color{orange}{-25}\cdot\color{red}{1} \]

  13. helder_edwin
    • 2 years ago
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    got this?

  14. helder_edwin
    • 2 years ago
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    http://en.wikipedia.org/wiki/Hyperbola

  15. HopelessMathStudent
    • 2 years ago
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    mhm :)

  16. helder_edwin
    • 2 years ago
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    now \[ \large 16(x-\color{red}{2})^2-25(y\color{red}{+1})^2=400 \] \[ \large \frac{16(x-\color{red}{2})^2}{400}-\frac{25(y-(\color{red}{-1}))^2}{400}=1 \] \[ \large \frac{(x-\color{red}{2})^2}{25}-\frac{(y-(\color{red}{-1}))^2}{16}=1 \]

  17. helder_edwin
    • 2 years ago
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    got this?

  18. HopelessMathStudent
    • 2 years ago
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    I see :)

  19. helder_edwin
    • 2 years ago
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    now the center of the hyperbola is \[ \large (\color{red}{2},\color{red}{-1}) \]

  20. helder_edwin
    • 2 years ago
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    is it a horizontal or vertical hyperbola? how can u tell?

  21. HopelessMathStudent
    • 2 years ago
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    horizontal

  22. HopelessMathStudent
    • 2 years ago
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    the format of the equation.

  23. helder_edwin
    • 2 years ago
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    no.

  24. helder_edwin
    • 2 years ago
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    look at the denominators which is larger? under x? or under y?

  25. HopelessMathStudent
    • 2 years ago
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    x.....

  26. helder_edwin
    • 2 years ago
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    yes. this tells u the hyperbola is horizontal

  27. HopelessMathStudent
    • 2 years ago
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    I said it was horizontal. Is it B?

  28. helder_edwin
    • 2 years ago
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    what is the length of the semi-mayor axis?

  29. helder_edwin
    • 2 years ago
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    it HAS to. do u know why?

  30. HopelessMathStudent
    • 2 years ago
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    because the formula is formatted (h±a,k) for horizontal and that fits? lol

  31. helder_edwin
    • 2 years ago
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    yes. very clever. u were looking for the foci not the vertices, though.

  32. helder_edwin
    • 2 years ago
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    did u understand?

  33. HopelessMathStudent
    • 2 years ago
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    do it is B? :) and yeah pretty much lol

  34. HopelessMathStudent
    • 2 years ago
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    so*

  35. helder_edwin
    • 2 years ago
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    i hope so.

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