## sloppycanada one year ago Ellipses http://gyazo.com/e897fc96e592ab1e5dcba825538b6d7e y^2/100 - x^2/16 = 1

1. anonymous

no not this time

2. anonymous

center is not $$(0,0)$$ find the center first

(1,2)?

4. anonymous

yes

5. anonymous

so it is going to look like $\frac{(x-1)^2}{a^2}-\frac{(y-2)^2}{b^2}=1$

6. anonymous

you need $$a$$ and $$b$$

But wouldn't x and y be switched? Since the graph is going a different way?

8. anonymous

no

So it's not like an ellipses that way.

10. anonymous

|dw:1435717476572:dw|

11. anonymous

scuse the crappy pictures

No worries, mine are worse Okay so for this graph it'd be - (x^2-2)/100 - (y^2-1)/16 = 1

Also for this one - you said it was x than y. http://gyazo.com/99da097eae7203d795aaab65586d4f48

14. anonymous

15. anonymous

also no, not 100 and 16

16. anonymous

the distance from the center to the vertex is 5

17. anonymous

not the whole length from one vertex to the other, which is 10 so $$25$$ goes under the $$x$$ part

18. anonymous

similarly under the $$y$$ part put a 4, not 16

(x^2-2)/25 - (y^2-1)/4 = 1 is my final answer?

20. anonymous

yes want to check it?

21. anonymous

oh no !!

No?

23. anonymous

not $$x^2-1$$ up top but rather $$(x-1)^2$$

24. anonymous

sorry $$(x-2)^2$$

(x-2)^2/25 - (y-1)^2/4 = 1

26. anonymous

here is the check that $\frac{(x-2)^2}{25}-\frac{(y-1)^2}{4}=1$ is right http://www.wolframalpha.com/input/?i=hyperbola+%28x-1%29^2%2F25-%28y-2%29^2%2F4%3D1

27. anonymous

hard to tell from the graphs, but you can see from the vertices that it is correct

Okay thanks!

So many hyperbola questions...

30. anonymous

there is a certain sameness to them

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