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sukhadia1 · Jun 2, 2011, 07:24 PM

4x^2+9y^2-24x-90y+225=0

jcaron2 · Jun 2, 2011, 08:10 PM

This is the equation of an ellipse, so let's first put it in standard form. The standard form is:

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

In order to get your equation in that form, we'll need to complete the square for the terms with x and y.

$4x^2+9y^2-24x-90y+225=0$

$(4x^2-24x)+(9y^2-90y)=-225$

$(4x^2-24x+36)+(9y^2-90y+225)=-225+36+225$

$(2x-6)^2+(3y-15)^2=36$

$4(x-3)^2+9(y-5)^2=36$

$\frac{(x-3)^2}{9}+\frac{(y-5)^2}{4}=1$

$\frac{(x-3)^2}{3^2}+\frac{(y-5)^2}{2^2}=1$

There! Now it's in standard form with h=3, k=5, a=3, and b=2. From that information can you find the center, vertices, and co-vertices? If you still don't know how to proceed, post back and one of us will guide you the rest of the way.

sukhadia1 · Jun 2, 2011, 08:40 PM

Yeah, that's exactly what I got I just was not so sure if it was right, I got the vertices (6,5), (0,5) because of the greater axis? If I'm right it's a horizontal parabola?

sukhadia1 · Jun 2, 2011, 08:41 PM

Yeah, that's exactly what I got I just was not so sure if it was right, I got the vertices (6,5), (0,5) because of the greater axis? If I'm right it's a horizontal parabola?

jcaron2 · Jun 3, 2011, 05:25 AM

Perfect! (Except I assume you meant to write horizontal ellipse, not horizontal parabola :) ).

The co-vertices, as I'm sure you know, are at (3,3) and (3,7).