# SOLUTION: Find the vertex, focus, directrix, and Axis of symetry for: x^2+10x+4y+9=0 I do not know what to do. :( I have been able to bring it to (x+5)^2 +4y =-9, but because there is no

Algebra ->  Algebra  -> Quadratic-relations-and-conic-sections -> SOLUTION: Find the vertex, focus, directrix, and Axis of symetry for: x^2+10x+4y+9=0 I do not know what to do. :( I have been able to bring it to (x+5)^2 +4y =-9, but because there is no       Log On

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 Question 606915: Find the vertex, focus, directrix, and Axis of symetry for: x^2+10x+4y+9=0 I do not know what to do. :( I have been able to bring it to (x+5)^2 +4y =-9, but because there is no Y^2, I don't know what to do. :(Answer by ewatrrr(10682)   (Show Source): You can put this solution on YOUR website! ``` Hi, Yes, there is no y^2. This is a Parabola. the vertex form of a parabola opening up or down, where(h,k) is the vertex. x^2+10x+4y+9=0 (x+5)^2 -25 + 4y + 9=0 (x+5)^2 + 4y -16 =0 a = -1/4<0, open downwards, Center(-5,4) Axis of symmetry x=-5 The standard form is , where the focus is (h,k + p) 4p = -4, p = -1 , focus is (-5,3) and directrix is y = 5 See below descriptions of various conics Standard Form of an Equation of a Circle is where Pt(h,k) is the center and r is the radius Standard Form of an Equation of an Ellipse is where Pt(h,k) is the center. a and b are the respective vertices distances from center and ±are the foci distances from center Standard Form of an Equation of an Hyperbola opening right and left is: where Pt(h,k) is a center with vertices 'a' units right and left of center. Standard Form of an Equation of an Hyperbola opening up and down is: where Pt(h,k) is a center with vertices 'b' units up and down from center. the vertex form of a parabola opening up or down, where(h,k) is the vertex. The standard form is , where the focus is (h,k + p) the vertex form of a parabola opening right or left, where(h,k) is the vertex. The standard form is , where the focus is (h +p,k ) ```