SOLUTION: Identify the conic section. If it is a parabola, give the vertex. If it is an ellipse or a hyperbola, give the center and the foci. {{{ y^2-4x+6y+29=0 }}}

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Identify the conic section. If it is a parabola, give the vertex. If it is an ellipse or a hyperbola, give the center and the foci. {{{ y^2-4x+6y+29=0 }}}      Log On


   

Question 621143: Identify the conic section. If it is a parabola, give the vertex. If it is an ellipse or a hyperbola, give the center and the foci.
+y%5E2-4x%2B6y%2B29=0+
Found 2 solutions by ewatrrr, lwsshak3:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

+y%5E2-4x%2B6y%2B29=0+

+%28y%2B3%29%5E2+%2B+20=+4x+

+%281%2F4%29%28y%2B3%29%5E2+%2B+5=+x+ Parbola opening to the Right V(5,-3) y = -3 line of symmetry

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
y^2-4x+6y+29=0
complete the square
(y^2+6y+9)-4x=-29+9
(y+3)^2=4x-20
(y+3)^2=4(x-5)
This is an equation of a parabola that opens rightwards.
Its standard form: (y-k)^2=4p(x-h), (h,k)=(x,y) coordinate of vertex.
For given equation:
vertex: (5,-3)
see graph below:
y=±3(4x-20)^.5-3