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? :/

Question

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? :/

campbell_st · Answer

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

anonymous · Answer

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

campbell_st · Answer

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

anonymous · Answer

ah ok..

anonymous · Answer

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 ?

anonymous · Answer

$$
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

 
$$