How do you identity if the equation #7x^2-28x+4y^2+8y=-4# is a parabola, circle, ellipse, or hyperbola and how do you graph it?

1 Answer
Dec 28, 2016

Answer:

It's an ellipse with #x#-semiaxis #2# and #y#-semiaxis #sqrt(7)# centred on the point #(2,-1)#

Explanation:

It is probably an ellipse because the coefficients of #x^2# and #y^2# have the same sign but different values, and there is no term in #xy#.

So complete the square for the #x# and #y# separately:
#7x^2-28x+4y^2+8y=-4#
#7(x-2)^2-28+4(y+1)^2-4=-4#
#7(x-2)^2+4(y+1)^2=28#
Then reduce to standard form:
#((x-2)/2)^2+((y+1)/sqrt(7))^2=1#
which is of the standard form:
#(x/a)^2+(y/b)^2=1#
so you can read off the semi-axes #a# and #b# and note that the centre is at #(2,-1)#.
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