anonymous one year ago What is the equation of the following graph?

1. anonymous

2. anonymous

@Michele_Laino

3. Michele_Laino

that is a traslated hyperbola

4. anonymous

|dw:1439232785706:dw| I have to use this equation right?

5. Michele_Laino

the requested equation is an equation like this: $\Large \frac{{{{\left( {x - {x_0}} \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {y - {y_0}} \right)}^2}}}{{{b^2}}} = 1$

6. Michele_Laino

from your graph we can see that: $\Large 2a = 10$

7. anonymous

wait, how do you know A is 2?

8. anonymous

isn't a "x" and b "y" ?

9. Michele_Laino

what is A?

10. Michele_Laino

referring to your drawing, we have: 2a=10, to understand that it is suffice to count the number of little squares

11. Michele_Laino

namely 2*a is the base of the dashed rectangle, whereas 2*b is the height of the dashed rectangle so 2*b=4

12. Michele_Laino

then we have: $\Large \begin{gathered} \frac{{{{\left( {x - {x_0}} \right)}^2}}}{{{5^2}}} - \frac{{{{\left( {y - {y_0}} \right)}^2}}}{{{2^2}}} = 1 \hfill \\ \hfill \\ \frac{{{{\left( {x - {x_0}} \right)}^2}}}{{25}} - \frac{{{{\left( {y - {y_0}} \right)}^2}}}{4} = 1 \hfill \\ \end{gathered}$

13. Michele_Laino

now, we establish what are x_0 and y_0

14. anonymous

im so confused ...

15. Michele_Laino

from your graph, I see that your hyperbola passes at point (6,2), so we have: $\Large \frac{{{{\left( {6 - {x_0}} \right)}^2}}}{{25}} - \frac{{{{\left( {2 - {y_0}} \right)}^2}}}{4} = 1$

16. anonymous

but how did you get the 25?

17. Michele_Laino

here is why:

18. Michele_Laino

19. anonymous

Okay, I get that! Now, do I have to simplify that equation, or is it left just like that?

20. Michele_Laino

as I said before, we have to establish the value of x_0 and y_0

21. anonymous

and how do we do that?

22. Michele_Laino

as we can see, the center of our dashed rectangle is located at point (1,2), right?

23. Michele_Laino

do you see that point?

24. anonymous

right!

25. Michele_Laino

in other words our hyperbola has been traslated by 1 unit to right and by 2 units to up

26. anonymous

okay, and what is the next step now?

27. Michele_Laino

that means x_0=1 and y_0=2, so, substituting them intoi my equation above, we get: $\Large \frac{{{{\left( {x - 1} \right)}^2}}}{{25}} - \frac{{{{\left( {y - 2} \right)}^2}}}{4} = 1$

28. Michele_Laino

into*

29. anonymous

so where did they 6 & 2 go?

30. Michele_Laino

x=6 and y=2 are the coordinates of a point which belongs to our hyperbola, infact, if we substitute x=6 and y=2 into my last formula, we get: $\large {\text{left side}} = \frac{{{{\left( {6 - 1} \right)}^2}}}{{25}} - \frac{{{{\left( {2 - 2} \right)}^2}}}{4} = \frac{{25}}{{25}} - 0 = 1 = {\text{right side}}$ namely I got an identity