## HopelessMathStudent 2 years ago What are the foci of the hyperbola given by the equation 16x2 - 25y2 - 64x - 50y - 361 = 0?

1. helder_edwin

complete the squares

2. HopelessMathStudent

How?

3. HopelessMathStudent

(-1 ± , 2) (2 ± , -1) (-1, 2 ± ) (2, -1 ± )

4. HopelessMathStudent

Its one of these lol Im just not sure which

5. jishan

|dw:1355058848205:dw|

6. helder_edwin

first factor the coeficientes of the squared terms (i re-arranged the terms) $\large 16x^2-64x-25y^2-50=361$ $\large 16(x^2-4x+\qquad)-25(y^2+2y+\qquad)=361$

7. helder_edwin

got this?

8. HopelessMathStudent

Kinda but what are the blank spaces?

9. helder_edwin

there's a "y" missing next to the 50.

10. HopelessMathStudent

oh okay

11. HopelessMathStudent

Pretty sure its A or B right?

12. 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 $\large \color{green}{16}(x^2-4x+\color{red}{4})\color{orange}{-25}(y^2+2y+\color{red}{1})=361 +\color{green}{16}\cdot\color{red}{4}\color{orange}{-25}\cdot\color{red}{1}$

13. helder_edwin

got this?

14. helder_edwin
15. HopelessMathStudent

mhm :)

16. helder_edwin

now $\large 16(x-\color{red}{2})^2-25(y\color{red}{+1})^2=400$ $\large \frac{16(x-\color{red}{2})^2}{400}-\frac{25(y-(\color{red}{-1}))^2}{400}=1$ $\large \frac{(x-\color{red}{2})^2}{25}-\frac{(y-(\color{red}{-1}))^2}{16}=1$

17. helder_edwin

got this?

18. HopelessMathStudent

I see :)

19. helder_edwin

now the center of the hyperbola is $\large (\color{red}{2},\color{red}{-1})$

20. helder_edwin

is it a horizontal or vertical hyperbola? how can u tell?

21. HopelessMathStudent

horizontal

22. HopelessMathStudent

the format of the equation.

23. helder_edwin

no.

24. helder_edwin

look at the denominators which is larger? under x? or under y?

25. HopelessMathStudent

x.....

26. helder_edwin

yes. this tells u the hyperbola is horizontal

27. HopelessMathStudent

I said it was horizontal. Is it B?

28. helder_edwin

what is the length of the semi-mayor axis?

29. helder_edwin

it HAS to. do u know why?

30. HopelessMathStudent

because the formula is formatted (h±a,k) for horizontal and that fits? lol

31. helder_edwin

yes. very clever. u were looking for the foci not the vertices, though.

32. helder_edwin

did u understand?

33. HopelessMathStudent

do it is B? :) and yeah pretty much lol

34. HopelessMathStudent

so*

35. helder_edwin

i hope so.

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