Posted by **Cal** on Sunday, April 19, 2009 at 6:33pm.

Find the standard form of the equation of the ellipse. (Remember center is midpoint between either foci or vertices)

(1) 9x^2+4y^2+36x-24y+36=0

(2) Vertices: (0,5)(0,-5)

Passes through the point (4,2)

Centered at the origin

- Pre-Cal -
**Reiny**, Sunday, April 19, 2009 at 7:31pm
for the first one, complete the square :

9x^2+4y^2+36x-24y+36 = 0

9(x^2 + 4x + ____) + 4(y^2 - 6y + ____) = -36

9(x^2 + 4x + 4) + 4(y^2 - 6y + 9) = -36+36+36

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

divide each term by 36

(x+2)^2/4 + (y-3)^2/9 = 1

for 2. remember that from the vertices we know a = 5

standard form with centre at (0,0) is

x^2/a^2 + y^2/b^2 = 1

so we have

x^2/25 + y^2/b^2 = 1

but (4,2) lies on it, so

16/25 + 4/b^2 = 1

I will leave it up to you to solve for b^2, and plug that back into the above equation.

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