# SOLUTION: Question: Find the equation for the parabola with focus at (-1,-1) and vertex at (4,-1) POssible Answers: (a) x^2-8x+20y+28=0 (b) -4x^2-7x-4y^2+5y-5+0 (c) 5y^2-7x-5y=-8 (d)

Algebra ->  Algebra  -> College  -> Linear Algebra -> SOLUTION: Question: Find the equation for the parabola with focus at (-1,-1) and vertex at (4,-1) POssible Answers: (a) x^2-8x+20y+28=0 (b) -4x^2-7x-4y^2+5y-5+0 (c) 5y^2-7x-5y=-8 (d)      Log On

 Algebra: Linear Algebra (NOT Linear Equations) Solvers Lessons Answers archive Quiz In Depth

 Question 30360: Question: Find the equation for the parabola with focus at (-1,-1) and vertex at (4,-1) POssible Answers: (a) x^2-8x+20y+28=0 (b) -4x^2-7x-4y^2+5y-5+0 (c) 5y^2-7x-5y=-8 (d)5x^2-7x+10y^2-5y=-3 Thanks!Answer by sdmmadam@yahoo.com(530)   (Show Source): You can put this solution on YOUR website!Find the equation for the parabola with focus at (-1,-1) and vertex at (4,-1) The focus S(-1,-1) and the vetex A(4,-1) are on the line y =-1 which is a horizontal line parallel to the x-axis at a distance 1 unit below it Axis horizontal means the parabola is square in y and linear in x The focus is to the left of the vertex, which means the parabola is facing west The equation to the parabola will be of the form (y-k)^2 = -4a(x-h)----(1) Where A(h,k ) = A(4,-1) and a = distance of the focus from the vertex. The distance SA = (one unit to the y-axis plus 4 units to the vertex) = 1+4 = 5 Therefore putting h =4, k = -1 and a = 5 Therefore (1) becomes [y-(-1)]^2 = -4X5(x-4) (y+1)^2 = -20(x-4) y^2+2y+1= -20x+80 y^2+2y+20x+1-80 =0 y^2+2y+20x-79 = 0 which DOES NOT tally with any of the following given answers: (a) x^2-8x+20y+28=0 (b) -4x^2-7x-4y^2+5y-5+0 (c) 5y^2-7x-5y=-8 (d)5x^2-7x+10y^2-5y=-3 Remark: Either the set of answers given is wrong or the given particulars about the parabola are wrong. Please give the correct problem