Find the standard form of the equation of the parabola with the given characteristic(s) and vertex at the origin. Focus: (-9,0) Focus: (0, 1/6) Directrix: 2

Answer 1

Focus (- 9, 0); y² = 4px, p = - 9; y² = - 36x ;

Focus (0, 1/6); x² = 4py, p = 1/6; x² = 2/3y

Directrix x = 2;then p = - 2 and parabola y² = - 8x

y = 2, then p = -2 and parabola x² = - 8y

Answer 2

for focus (0,-9), the parabola is y=-x^2/36

for focus (0,1/6) the parabola equation is y=3x^2/2

for directrix = 2 it's y=-x^2/8

only difference is the slope

all have vertices at the origin, so (y-h) = a(x-k)^2 becomes y=ax^2. All we need is to find a, the coefficient of the x squared term. they are -1/36, 3, and -1/8

for each parabola find a point on the parabola, to the right of the focus. It's distance horizontally to the vertex = the distance from the vertex to the directrix. For focus -9, that point on the parabola is (18,-9) plug it in and solve for a = -1/36. Same method for the other focus and the directrix.