Find the equation in standard form of the parabola

Question

1.Find the equation in standard form of the parabola with focus (2, 1)

 and directrix x = −8.

2.Find the equation in standard form of the parabola that has vertex (8, −4),

 has its axis of symmetry parallel to the x-axis, and passes through the point (5, 1).

3.Find the equation in standard form of the parabola with focus (6, 9)

 and directrix y = 7

4.Find the equation in standard form of the parabola with vertex at (9, 3)

 and focus (9, 0).

Answer 1

1) 20x = (y-1)^2 -3

2) 196x = -3(y+4)^2 +8

3) 4y = (x-6)^2 + 8

4) -12y = (x-9)^2 + 3

1) vertex is half way between focus and directrix V(-3,1) x=a(y-1)^2 -3

to solve for a plug in another point on the parabola (2,11), a point directly above the focus, and equidistant to the directrix, 10 from 2 to -8, 10 from (2,1) to (2,11)

2 Vertex (8-4) with another point (5,11) to the right and above, if it's a lefttward opening parabola, then it's x=a(y+4) + 8. solve for a by plugging in x=5, y=11. a should be <0.

but if it's an upward opening parabola then y=a(x-8)^2 -4, solve for a by plugging in the point (5,11) a should be positive: 3y=5(x-8)^2 -4

the parabola could be either one, leftward or upward opening

3 It's an upward opening parabola, since focus is above the directrix. vertex is halfway between (6,8)

y=a(x-6)^2 + 8, find another point on the parabola and plug it in to solve for a a should be positive

find the point directly left of the focus (4,9) equidistant to the directrix. a =4.

1. vertex = (9,3) focus is below, so it's a downward opening parabola. y=a(x-9)^2 +3. plug in another point to solve for a. It should be negative for "a" find the point (3,0) directly left of the focus equidistant to the diretrix which is y=6