What is the equation of the parabola shown below, given a focus at F(−2, 5) and a directrix of y = −1? In addition, identify the vertex and the equation of the axis of symmetry of the parabola.

What is the equation of the parabola shown below given a focus at F2 5 and a directrix of y 1 In addition identify the vertex and the equation of the axis of sy class=

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Answer:

The equation of the parabola is:

[tex]y=\frac{1}{12}(x+2)^2+2[/tex]

The vertex, (h, k) = (-2, 2)

The equation of the axis of symmetry is x = -2

Option A is the correct choice

Explanation:

The focus, (h, f) = (-2, 5)

That is, h = -2, f = 5

The directrix, y = -1

The distance from the focus to thevertex = f - k

The distance from the vertex to the directrix = k - (-1)

The distance from the vertex to the directrix = k + 1

f - k = k + 1

Since f = 5

5 - k = k + 1

k + k = 5 - 1

2k = 4

k = 2

The vertex, (h, k) = (-2, 2)

The equation of the parabola is of the form:

y = a(x - h)² + k

[tex]\begin{gathered} \text{where a = }\frac{1}{4(f-k)} \\ a=\frac{1}{4(5-2)} \\ a=\frac{1}{12} \end{gathered}[/tex]

Substititute a = 1/12, h = -2, and k = 2 into the equation y = a(x - h)² + k

[tex]\begin{gathered} y=\frac{1}{12}(x-(-2))^2+2 \\ y=\frac{1}{12}(x+2)^2+2 \end{gathered}[/tex]

The equation of the parabola is:

[tex]y=\frac{1}{12}(x+2)^2+2[/tex]

The axis of symmetry of the parabola is the equation of the x-axis of the vertex

x = h

x = -2

The equation of the axis of symmetry is x = -2

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