# SOLUTION: Complete the square, then graph and identify the vertex, focus, directrix, and endpoints of the latus rectum for the equation -14x+2y^2-8y=20

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 Question 615731: Complete the square, then graph and identify the vertex, focus, directrix, and endpoints of the latus rectum for the equation -14x+2y^2-8y=20Answer by lwsshak3(6494)   (Show Source): You can put this solution on YOUR website!Complete the square, then graph and identify the vertex, focus, directrix, and endpoints of the latus rectum for the equation -14x+2y^2-8y=20 -14x+2y^2-8y=20 complete the square 2(y^2-4y+4)=20+14x+8 2(y-2)^2=14x+28 divide by 2 (y-2)^2=7x+14 (y-2)^2=7(x+2) This is an equation of a parabola that opens rightwards. Its standard form: (y-k)^2=4px, (h,k)=(x,y) coordinates of the vertex For given equation:(y-2)^2=7(x+2) vertex:(-2,2) axis of symmetry: y=2 4p=7 p=7/4 Focus: (-2+p,2)=(-2+7/4,2)=(-1/4,2) (p distance to the right of the vertex on the axis of symmetry) Directrix: x=(-2-p)=(-2-7/4)=-15/4 (p distance to the left of the vertex on the axis of symmetry) latus rectum: length of latus rectum=4p=7 2p=7/2 end points: (-1/4,2±2p) =(-1/4,2±7/2) =(-1/4,-1.5) and (-1/4,5.5) see graph below: y=(7(x+2))^.5+2