SOLUTION: Find the center, foci, vertices, asymptotes, and radius, as appropriate, of the conic section: {{{ 2x^2 + 2y^2 - 28x + 12y + 114 = 0 }}} ----------- ----------- ----------- --

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the center, foci, vertices, asymptotes, and radius, as appropriate, of the conic section: {{{ 2x^2 + 2y^2 - 28x + 12y + 114 = 0 }}} ----------- ----------- ----------- --      Log On


   



Question 1032842: Find the center, foci, vertices, asymptotes, and radius, as appropriate,
of the conic section:
+2x%5E2+%2B+2y%5E2+-+28x+%2B+12y+%2B+114+=+0+
-------------------------------------------
Here is what I did...I am just not sure if I did it correctly.
+2x%5E2+%2B+2y%5E2+-+28x+%2B+12y+%2B+114+=+0+
=+2x%5E2+-+28x+%2B+2y%5E2+%2B+12y++=+-114+
=+2%28x%5E2+-+14x%29+%2B+2y%5E2+%2B+12y++=+-114+
=+2%28x%5E2+-+14x+%2B+49%29+%2B+2y%5E2+%2B+12y++=+-114+%2B+98+
=+2%28x+-+7%29%5E2+%2B+2y%5E2+%2B+12y++=+-114+%2B+98+
=+2%28x+-+7%29%5E2+%2B+2%28y%5E2+%2B+6y%29++=+-114+%2B+98+
=+2%28x+-+7%29%5E2+%2B+2%28y%5E2+%2B+6y+%2B+9%29++=+-114+%2B+98+%2B+18+
=+2%28x+-+7%29%5E2+%2B+2%28y+%2B+3%29%5E2++=+2+
=+%28x+-+7%29%5E2+%2B+%28y+%2B+3%29%5E2++=+1+
Center: (7, -3)
Foci: (7, +sqrt%283%29+ - 3)
Vertices: (7, -2)
Asymptotes: N/A
Radius: N/A
Did I do this right? I just have the gut feeling that I messed up somewhere.
Thank you!

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Find the center, foci, vertices, asymptotes, and radius, as appropriate,
of the conic section:
============
+2x%5E2+%2B+2y%5E2+-+28x+%2B+12y+%2B+114+=+0+
Divide by 2, make it easier.
x%5E2+%2B+y%5E2+-+14x+%2B+6y+%2B+57+=+0+
x%5E2+%2B+y%5E2+-+14x+%2B+6y+=+-57+
x%5E2-14x++%2B+y%5E2%2B6y+=+-57
x%5E2-14x%2B49++%2B+y%5E2%2B6y%2B9+=+-57+%2B+58+=+1
%28x-7%29%5E2+%2B+%28y%2B3%29%5E2+=+1
--> circle of radius 1, center at (7,-3)