Easy speasy but i dont get it.

[shimmeringlight]

okay, so math

it says...

Determine an equation in standard from for each hyperbola...
centre (-2,1), one focus (-4,1), length of conjugate axis 4


what i figured out... but its wrong... because i can't figure out the answer!

(x+2)^2 - (y-1)^2 = 1
a^2_____[/color]b^2

conjugate axis is 4 so co-ordinates for the co-vertices or vertices should be
(-2,3), (-2,-1)

so then to find find the a or b value... i use
c^2=a^2+b^2
2^2=a^2+2^2

but 4-4=0... and that's impossible? i got stuck here...

help please... thank you in advance.

 

[moose]

c^2=a^2+b^2

I always thought that in these problems, you would use c^2=a^2-b^2...Could someone check that?

 

[shimmeringlight]

I always thought that in these problems, you would use c^2=a^2-b^2...Could someone check that?

i checked it but my book says c^2=a^2+b^2 unless the book is wrong... but I've done other problems and its worked out... I am just stuck on this one question.

 

[Ouabache]

\frac {(x+2)^2}{a^2}- \frac{(y-1)^2}{b^2} =1 your first equation looks fine.

I needed to review conjuage axis as well as transverse axis as it pertains to hyperbolae.
You may want to review that page too, as it may offer some clues on how to solve this problem.
They also confirm your relation c^2=a^2+b^2

Another good hyperbola reference from mathworld.

 

[theCandyman]

Moose is corret, c^2 = a^2 - b^2. You will have to post the orginal problem for me to help you anymore, I do not really remember ellipses well.

For reference, standard form: \frac{x-h^2}{a^2} - \frac{y-k^2}{b^2} = 1, where the ellipse is centered at (h, k).

 

[shimmeringlight]

ok, i figured it out. thx

 

[Ouabache]

Moose is corret, c^2 = a^2 - b^2. You will have to post the orginal problem for me to help you anymore, I do not really remember ellipses well.

That equation is great for an ellipse, this is a hyperbola question.

 

[theCandyman]

That equation is great for an ellipse, this is a hyperbola question.

Hahaha

I was wondering why I could not do much of the problem, I think mentioning two vertices threw me off the hyperbola, I always associate them with ellipses.