find the equation of hyperbola. conjugate axis on the x-axis, one focus at (0, √13), equation of one directrix y= 9√13 / 13. HELP PLEASE!!!!!!! the answer is 4y^2 - 9x^2 = 36... it was stated on the book

- anonymous

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- anonymous

Sure, do you have any idea on how to start or what the equation of a hyperbola looks like in general?

- anonymous

(x/a)^2 - (y/b)^2 = 1

- anonymous

nah. no idea.. sorry

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## More answers

- anonymous

Yes that's right!

- anonymous

the equation is correct! (:

- anonymous

Now, we need to use the given information to deduce what a and b are.

- anonymous

okay okay.. :)

- anonymous

We have one of the focus, so we can graph this, usually not necessary though, to see in which direction the hyperbola opens up.

- anonymous

sqrt(13)= 3.605

- anonymous

|dw:1378530253315:dw|

- anonymous

So could you graph the point (0, sqrt(13))?

- anonymous

|dw:1378627196153:dw|

- anonymous

okay okay.. :)

- anonymous

??

- anonymous

|dw:1378530449815:dw|

- anonymous

yes that's what i did.. hehe.

- anonymous

Right, its just a point not a line.

- anonymous

yes.. sorry. hehe

- anonymous

|dw:1378530601557:dw|

- anonymous

Okay, try again. Remember sqrt(13)= 3.605
I graphed the conjugate axis for ya.
Its the line with the pencil little bit curved as you can see, I couldn't make it a different color cus that tool is not available but anyways go ahead and tell me where the focus (0,sqrt(13)) is.

- anonymous

its between 3 and 4, in the y axis..

- anonymous

|dw:1378530971954:dw|

- anonymous

yes right there! (:

- anonymous

So with this focus we know the hyperbola opens up and down correct?

- anonymous

They say that the equation of the directrix is y = 0sqrt(13)/13
So where would this be?

- anonymous

@silverxx

- anonymous

the directrix is between 2 and 3, y axis.. ( y= 9√13 / 13 = 2.5)

- anonymous

That's right!|dw:1378531276923:dw|

- anonymous

That's the directrix! (:

- anonymous

So the vertex of that first hyperbola should be half way between the focus and the directrix, correct?

- anonymous

??

- anonymous

yes yes..

- anonymous

So where would the vertex be?

- anonymous

(0,0)?

- anonymous

so said yes, before to it so I thought you were sure about it?

- anonymous

yes im sure..

- anonymous

I said that the vertex would be between the directrix and the focus. Half ways between directriz.
W already found the directrix and the focus. So what would be the vertex?

- anonymous

|dw:1378531735674:dw|

- anonymous

near to three..

- anonymous

its actually 3

- anonymous

|dw:1378532185640:dw|

- anonymous

Remember that equation is (y/a)^2 - (x/b^2=1
Does that make sense?

- anonymous

|dw:1378532371544:dw|

- anonymous

Remember that equation is (y/a)^2 - (x/b^2=1
Does that make sense?

- anonymous

yes...

- anonymous

|dw:1378532439536:dw|

- anonymous

We are trying to find the equation, we already have a is equal to, now we need b.
Do you see that equations I put in a polygon? That's the equation we need. You gave it to me at the beginning. But we need to find the values of a and b. Correct? No we need the value for b.

- anonymous

|dw:1378532633022:dw|

- anonymous

Now any ideas how to find b?

- anonymous

x_x

- anonymous

its 2?

- anonymous

That's right!

- anonymous

Do you know how to find it?
There is a formula for it

- anonymous

id just observed it in the graph.. maybe the pythagorean but c's not given..

- anonymous

|dw:1378532974473:dw|
Where p is the distance from the focus to the directrix, and c is the distance from the center to the focus. We know that the center is (0,0) and the focus is (0, sqrt(13)
So we can calculate all these distances. But yes, b=2. So now you can plug into the equation and you have it all done. (:

- anonymous

Have a good day, and good luck with your studies! (:
I hope after my explanation- you've understood this clearly!

- anonymous

thank you so much.. im so sorry for being so.. well, you know. thank you thank you!!!

- anonymous

haha, your welcome! (:

- anonymous

how did u come up the 3 and 2 ?sorry...

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