# Write the equation of the parabola whose directrix y=-9.25 and has a focus at (9,-6.75)?

• Write the equation of the parabola whose directrix y=-9.25 and has a focus at (9,-6.75)?

Answer #1 | 04/03 2017 20:19
Directrix is a horizontal line so the parabola is of the form (x-h)^2 = 4p(y-k) where (h,k) is the vertex coordinates of the focus is (h,k+p) h=9 k+p = -6.75 ------(1) Equation of the dirrectrix is y= k-p y = -9.25 k-p = -9.25 ------(2) add (1) & (2) 2k = -16 k=-8 vertex is (h,k) = (9,-8) (x-9)^2 = 4p (y-(-8)) (x-9)^2 = 4p (y+8) --- equation of the parabola k+p = -6.75 -8+p = -6.75 p = 8-6.75 = 1.25 (x-9)^2 = (4)(1.25)(y+8) (x-9)^2 = 5(y+8) (1/5) (x-9)^2 = y+8 y = (1/5)(x-9)^2 - 8
Positive: 50 %
Answer #2 | 04/03 2017 21:00
y = -9.25 is a horizontal line. A horizontal directrix means a vertical parabola. The focus lies above the directrix, so the parabola opens upwards. The vertex is halfway between focus and directrix, at (9, -8). General equation for up-opening parabola:  y = a(x-h)² + k with  a > 0  vertex (h,k)  focal length p = 1/(4a)  focus (h,k+p)  directrix y = k-p Apply your data and solve for h, k, and a vertex (h,k) = (9,-8) h = 9 k = -8 p = distance between focus and vertex = 1.25 a = 1/(4p) = ⅕ The equation becomes y = ⅕(x-9)² - 8
Positive: 33.333333333333 %
Answer #3 | 04/03 2017 20:31
Directrix: y = −9.25 Focus = (9, −6.75) Vertex = (9, (−6.75−9.25)/2) = (9, −8) p = −6.75−(−8) = 1.25 y = 1/(4p) (x−h)² + k, where (h,k) = vertex y = 1/5 (x−9)² − 8
Positive: 0 %
Answer #4 | 04/03 2017 20:05
Let (x,y) be on the parabola. Distance squared to focus (9, -6.75) is (x-9)^2 + (y+6.75)^2 Distance squared to directorix is (y + 9.25)^2 Those two expressions are equal. Expanding the squared terms with y we have: (x - 9)^2 + y^2 + 13.5y + 45.5625 = y^2 + 18.5y + 85.5625 The y^2 terms cancel out, leaving (x - 9)^2 = 5y + 40 0.2(x - 9)^2 = y + 8 y = 0.2(x - 9)^2 - 8 is the vertex form for the parabola.
Positive: 0 %

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