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Mathematics 61 Online
OpenStudy (anonymous):

Complete the square to determine whether the equation represents an ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an ellipse, find the center, foci, vertices, and lengths of the major and minor axes. If it is a parabola, find the vertex, focus, and directrix. If it is a hyperbola, find the center, foci, vertices, and asymptotes. (If an answer does not exist, enter DNE.) 9x^2 − 36x + 4y^2 = 0 Can somebody help me with this? :/

OpenStudy (campbell_st):

9x^2 - 36x + 36 + 4y^2 = 36 (3x - 6)^2 + (2y)^2 = 36

OpenStudy (anonymous):

one question when you completed the square how come it is 9x^2-36x+36?

OpenStudy (campbell_st):

the middle term is -36x and for a perfect square the middle term of (a +b)^2 is 2ab so if a = 3x then 2*3x*b = -36x or 6xb = -36x divide both by 6x gives b = -6

OpenStudy (anonymous):

ah ok..

OpenStudy (anonymous):

ok so now i have 9(x-2)^2+4y=4 in order to make it an ellipse do I change it to [9(x-2)^2]/4+y^2=1 ?

OpenStudy (anonymous):

\[ 9 x^2 -36 x + 4 y^2 \\ 9 x^2 - 36 x + 36 -36 + 4 y^2 =0 \\ \ 9( x^2 -4 x + 4) + 4 y^2 = 36\\ 9( x-2)^2 + 4 y^2 = 36\\ \frac {( x-2)^2}{4}+ \frac{y^2}{9}=1 \]

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