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mollyjo614 · May 13, 2009, 08:42 PM

So I have the equation y^2/9 - x^2/36=1 how do I find the foci and vertices?

Perito · May 14, 2009, 07:11 AM

y^2/9 - x^2/36=1
How do I find the foci and vertices?

I'll assume this is what you meant:

$\frac {y^2}{9} - \frac {x^2}{36}=1$

The general form of an equation for an ellipse is

$\frac {(x-h)^2}{a^2} + \frac {(y-k)^2}{b^2} = 1$

Equation of Ellipse

Your equation is almost in this form:

$\frac {(x-0)^2}{3^2} + \frac {(y-0)^2}{6^2} = 1$

The center is at (h,k).

See this to determine the foci

Focus of Ellipse. The formula for the focus and ...

mznita15 · May 18, 2012, 09:19 AM

(x+2)^2 (y+1)^2
______ - _______ = 1 I'm looking for vertices,foci, and center
169 4

eulalio · Nov 21, 2012, 04:34 PM

Find the equation for the hyperbola that satisfies the given conditions? For Vertices(0, + or - 7) asymptote y= +or - 1/4.

eulalio · Nov 21, 2012, 04:36 PM

Find the equation for the hyperbola that satisfies the given conditions? For Vertices(0, + or - 7) asymptote y= +or - 1/4.

My answer was wrong y^2/49-16x^2/49=1
elp to see where I went wrong with this equation.