The equation y^2+4y+4x^2+4x+21=0 can be changed into a vertex equation by completing the square. Which conic section can be made from the graph of this equation?

A. ellipse B. hyperbola C. circle D. parabola

Question

The equation y^2+4y+4x^2+4x+21=0 can be changed into a vertex equation by completing the square. Which conic section can be made from the graph of this equation? 

A. ellipse B. hyperbola C. circle D. parabola

anonymous · Answer

there is $y^2$ and also $4x^2$ both with plus signs
if it the coefficients were the same, it would be a circle
but because they are different, it is an ellipse, a sort of flattened circle

anonymous · Answer

so ellipse is a circle just a flat one? how does that work?

anonymous · Answer

hold on one second, let me check something

anonymous · Answer

ok

anonymous · Answer

very strange
is there a typo in the question, or is it really $$y^2+4y+4x^2+4x+21=0$$

anonymous · Answer

they want you to say it is an ellipse, but in fact it is nothing
if you try to complete the square  you see there is no solution to this
very odd

anonymous · Answer

yes thats it what you just typed.

anonymous · Answer

thats exactly why im lost

anonymous · Answer

ok
what can i do?  it is nothing at all
but they want you to say it is an ellipse

anonymous · Answer

ok well you have helped me enough i guess then thanks

anonymous · Answer

yw