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

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 (http://www.analyzemath.com/EllipseEq/EllipseEq.html)

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 ... (http://www.mathwarehouse.com/ellipse/focus-of-ellipse.php)

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

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

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.