SOLUTION: Hi, I have to identify the conic of the equation below and find the center, vertices, and asymptotes:
{{{x^2/9 + y^2/4 = 1}}}
I can't find any examples like this so I'm lost
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Question 372311: Hi, I have to identify the conic of the equation below and find the center, vertices, and asymptotes:
I can't find any examples like this so I'm lost here. Any help you can provide is greatly appreciated :)
Found 2 solutions by robertb, solver91311:
Answer by robertb(5830) (Show Source): You can put this solution on YOUR website!
It's an ellipse, with center at the origin (0,0), semi-major axis along the x-axis with length 3, and semi-minor axis 2. Vertices are (3,0) and (-3,0), and covertices (0,2) and (0,-2). There are no asymptotes of any kind (horizontal, vertical, slant, etc.)
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
^2}{a^2} \pm \frac{(y\ -\ k)^2}{b^2}\ =\ 1)
If you have a plus sign between the LHS terms, and...
If 
, you have a circle center at )
, radius 
, but if 
you have an ellipse, center at , semi-major axis and semi-minor axis 
(unless 
and it is the other way around).
If you have a minus sign, then you have a hyperbola, centered at )
If only one of the variables is squared, then you have a parabola.
Only hyperbolas have asymptotes.
For your example, 
, 
, and 
. So you have an ellipse, centered at the origin, semi-major axis of 3, and semi-minor axis of 2. The vertices are at )
and )
, the end points of the semi-minor axis are at )
and )
. Calculate 
. Then the foci are at )
and )
. Ellipses do not have asymptotes.
John
My calculator said it, I believe it, that settles it
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