SOLUTION: Identify the conic section represented by each equation. For a poraola, give the vertex. For a circle, give the center and radius. For an ellipse or hyperbola, give the center and

Question 196696: Identify the conic section represented by each equation. For a poraola, give the vertex. For a circle, give the center and radius. For an ellipse or hyperbola, give the center and foci.
(1) x^2-4y^2+4x+8y-16=0
For this one I found that the equation would be (x+2)^2/16-(y-1)^2/4=1
I got that this is a hyperbola.
I found the center to be (-2,1)
I found the foci to be (4.47,0) (-4.47,0).

(2) 25x^2+4y^2-50x-24y-39=0
For this one I found that the equation would be (x-1)^2/4 + (y-3)^2/25=1
I got that this is an ellipse
I found that the center is (1,3)
I found that the foci is (0,4.58) (0,-4.58)
Is this correct?
I'm not really sure what I'm doing....
Thank you!!
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
(1) x^2-4y^2+4x+8y-16=0
For this one I found that the equation would be (x+2)^2/16-(y-1)^2/4=1
I got that this is a hyperbola.
I found the center to be (-2,1)
I found the foci to be (4.47,0) (-4.47,0)
The center is correct, but the foci can't be symmetrical about the Origin if the center is not at the Origin.
You have to add the "offset" to the foci, that is, shift them in the same way the center is shifted from the Origin.
foci at: (2.47,1) (-6.47,1)
---------------------------------
(2) 25x^2+4y^2-50x-24y-39=0
For this one I found that the equation would be (x-1)^2/4 + (y-3)^2/25=1
I got that this is an ellipse
I found that the center is (1,3)
I found that the foci is (0,4.58) (0,-4.58)
--------------
Same as above.
You have to add the "offset" to the foci, that is, shift them in the same way the center is shifted from the Origin.
--> foci at (1,7.58) (1,-1.58)
--------------
PS You can dl FREE software to graph these at
www.padowan.dk.com/graph/
Use Function, Insert Relation for these
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