PreCal
posted by Cal .
Find the standard form of the equation of the ellipse. (Remember center is midpoint between either foci or vertices)
(1) 9x^2+4y^2+36x24y+36=0
(2) Vertices: (0,5)(0,5)
Passes through the point (4,2)
Centered at the origin

for the first one, complete the square :
9x^2+4y^2+36x24y+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(y3)^2 = 36
divide each term by 36
(x+2)^2/4 + (y3)^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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