## sarah1017 one year ago What is the standard form of the equation of an ellipse with foci at (+-3,0) and co-vertices at (0,+-5)? Please show all steps!

1. ganeshie8

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2. ganeshie8

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3. ganeshie8

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4. ganeshie8

b = 5 a = ?

5. ganeshie8

if u can find a, u can write the equation of ellipse :- $$\large \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

6. ganeshie8

any idea how to find "a" ? :)

7. sarah1017

How do I find a? Is there a formula I have to follow?

8. ganeshie8

u just need to knw how to find an unknown side in a right triangle

9. sleung

Use the pythagorean theorem.

10. ganeshie8

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11. ganeshie8

a^2 = b^2 + c^2 a^2 = 5^2 + 3^2

12. ganeshie8

simplify

13. sarah1017

a^2=34

14. ganeshie8

yes, you want "a" right ?

15. sarah1017

Yes. So do I divide 34 by 2?

16. ganeshie8

noo, you need to take "square ROOT" both sides

17. sleung

So $a=\sqrt{34}$

18. sarah1017

5.83

19. ganeshie8

actually im wrong, a^2 is enough as u put it earlier,,,, i see we dont need to find a -.-

20. ganeshie8

directly plugin a^2 value in the equation

21. ganeshie8

$$\large \frac{x^2}{34} + \frac{y^2}{25} = 1$$

22. ganeshie8

^^final equation of ellipse

23. sarah1017

Whoaaa that confused me. So that's the final equation, is the work we did before still correct? Well except the finding a parts?

24. ganeshie8

everything is correct, you could even find "a", and plugin the "a" value back into the equaiton

25. ganeshie8

but thats just extra unnecessary work

26. ganeshie8

when u knw that u will be needing oly "a^2", why to find "a" ?

27. ganeshie8

it wont be a mistake to find "a", its just pain to take square root of 34 lol.. .hope u get me :)

28. sarah1017

I wasn't too sure how to do this at first so I just went with ya lol. I get it though. Thanks for your help :)

29. ganeshie8

np :) u wlc :))

30. sarah1017

:-)