## anonymous one year ago I need help with this question about hyperbolas and ellipses. The problem in in the attached file.

1. anonymous

2. jtvatsim

Ooh... this just looks like too much fun! :) You should probably start by drawing a picture of the situation for yourself. Also, Do you happen to have the formulas for the foci of hyperbolas and ellipses?

3. anonymous

|dw:1437111744452:dw|

4. anonymous

is that what it would look like?

5. anonymous

and no, I don't have the formulas for those

6. jtvatsim

Basically, the rectangle would just be touching the hyperbola, and the ellipse would fit perfectly inside, but I know drawing here is difficult. :)

7. jtvatsim

Let me find those equations for the foci.

8. anonymous

ok

9. jtvatsim

Focus of hyperbola: $a^2 + b^2 = f^2$ Focus of ellipse: $a^2 - b^2 = f^2$

10. jtvatsim

Let's start with the focus of the hyperbola since that's what your first bullet point is. Using the questions terminology we have $a^2 + b^2 = c_h^2$ which is just $c_h = \sqrt{a^2 + b^2}$ OK so far?

11. anonymous

ok

12. jtvatsim

Great! We can do the same thing for the ellipse. What equation should we start with?

13. anonymous

$a^2 + b^2 = c_e^2$

14. jtvatsim

Good! Just make that + into a - sign (it's a different equation for the ellipse) :)

15. anonymous

oh, that's right i forgot

16. jtvatsim

And I made a mistake by saying, we can do the "same thing", not exactly the same I guess. :)

17. jtvatsim

In any case, you will get $c_e = \sqrt{a^2 - b^2}$ Yay, we are getting closer!

18. anonymous

ok, great, so how do I started finding the distance between the two?

19. jtvatsim

Good question. That seems a little tough at first, but let's see if we draw the situation.

20. anonymous

ok

21. jtvatsim

|dw:1437112353871:dw|

22. jtvatsim

Well, now I can see that the distance between them is easy to find!

23. anonymous

i'm still confused on how to do that

24. jtvatsim

Well, for one thing, let's only talk about the positive focuses (foci). So just look at the right hand side.

25. anonymous

ok

26. jtvatsim

You can see that c_h is sitting far to the right, while c_e is sitting closer to the middle.

27. anonymous

yes

28. jtvatsim

Just to get you thinking... maybe c_h = 5 and c_e = 3. What would be the distance between them if that was true?

29. anonymous

2?

30. jtvatsim

Exactly! I would need to walk 2 units to get from one to the other. Now you found that by doing 5 - 3, right? You could have done c_h - c_e then to get the distance. But we have formulas for c_h and c_e! So all we do is subtract them and we get the distance.

31. anonymous

Oh, okay. So it would be $\sqrt{a^2+b^2}-\sqrt{a^2-b^2}$

32. jtvatsim

That right! And it makes sense too. a^2 + b^2 will give a bigger square root than a^2 - b^2 so our subtraction won't be negative or something weird.

33. anonymous

oh, that actually wasn't too difficult! Thank you so much

34. jtvatsim

That's always a great feeling when something crazy turns out to be easier than you thought. Your welcome! Good luck!

35. anonymous

Yeah, thanks!