Question

PLS HELP!!!!Pre-Cal
1. Find the equation of a parabola with vertex (2, -1), opens upward and has focal width of 16.
2. Find the equation of a parabola with focus (2, -3) and directrix x = 5.

Answer 1

Answer:

1. y = 1/16(x -2)^2 -1
2. x = -1/12(y +3)^2 -2

Step-by-step explanation:

1. For a parabola with vertex (h, k) that opens upward, the equation can be written ...

  y = 1/(4p)(x -h)^2 +k

where 4p is the focal width. Here, that is 16, and (h, k) = (2, -1), so the equation is ...

  y = 1/16(x -2)^2 -1

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2. The focal width is 4 times the distance from the focus to the directrix, so for this one we have ...

  4p = 4(2 -5) = -12

Because the parabola opens sideways, the roles of x and y in the equation are swapped. The equation for this parabola is ...

  x = 1/(4p)(y -k)^2 +h . . . . . . . for vertex (h, k) and focal width 4p

  x = -1/12(y +3)^2 +2

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The graph shows the two parabolas and their equations. The dashed lines are the focal width and the directrix.