# SOLUTION: Identify the conic section. If it is a parabola, give the vertex. If it is an ellipse or hyperbola, give the center and the foci. {{{ 8x^2-6y^2+48x-24y+0=0 }}}

Algebra ->  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 hyperbola, give the center and the foci. {{{ 8x^2-6y^2+48x-24y+0=0 }}}      Log On

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 Algebra: Conic sections - ellipse, parabola, hyperbola Solvers Lessons Answers archive Quiz In Depth

 Click here to see ALL problems on Quadratic-relations-and-conic-sections Question 620936: Identify the conic section. If it is a parabola, give the vertex. If it is an ellipse or hyperbola, give the center and the foci. Answer by lwsshak3(6522)   (Show Source): You can put this solution on YOUR website!Identify the conic section. If it is a parabola, give the vertex. If it is an ellipse or hyperbola, give the center and the foci. 8x^2-6y^2+48x-24y+0=0 8x^2+48x-6y^2-24y+0=0 complete the square 8(x^2+6x+9)-6(y^2+4y+4)=72-24 8(x+3)^2-6(y+2)^2=48 (x+3)^2/6-(y+2)^2/8=1 This is an equation of a hyperbola with horizontal transverse axis Its standard form: (x-h)^2/a^2-(y-k)^2/b^2=1, (h,k)=(x,y) coordinates of center For given hyperbola: center: (-3,-2) a^2=6 b^2=8 c^2=a^2+b^2=6+8=14 c=√14≈3.7 foci: (-3±c,-2)=(-3±3.7,-2)=(-6.7,-2) and (0.7,-2)