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Find the focus and directrix of the following parabola:

[tex](x-1)^2 = 8(y+4)[/tex]

Focus: ([tex]?, \square[/tex])

Directrix: [tex]y = \square[/tex]


Sagot :

Let's determine the focus and directrix of the given parabola equation:

[tex]\[ (x-1)^2 = 8(y+4) \][/tex]

The standard form of a parabola that opens vertically is:

[tex]\[ (x - h)^2 = 4p(y - k) \][/tex]

where [tex]\((h, k)\)[/tex] is the vertex of the parabola and [tex]\(p\)[/tex] is the distance from the vertex to the focus (or from the vertex to the directrix).

By comparing the given equation [tex]\((x - 1)^2 = 8(y + 4)\)[/tex] with the standard form [tex]\((x - h)^2 = 4p(y - k)\)[/tex], we can identify the following parameters:

- [tex]\( h = 1 \)[/tex]
- [tex]\( k = -4 \)[/tex]
- The coefficient [tex]\( 4p = 8 \)[/tex], thus [tex]\( p = 2 \)[/tex].

### Focus of the Parabola
The coordinates of the focus for a parabola that opens upwards or downwards are given by:

[tex]\[ (h, k + p) \][/tex]

Given that [tex]\( h = 1 \)[/tex], [tex]\( k = -4 \)[/tex], and [tex]\( p = 2 \)[/tex], we substitute these values in:

[tex]\[ \text{Focus} = (1, -4 + 2) = (1, -2) \][/tex]

### Directrix of the Parabola
The equation of the directrix for a parabola that opens upwards or downwards is:

[tex]\[ y = k - p \][/tex]

Substituting the values [tex]\( k = -4 \)[/tex] and [tex]\( p = 2 \)[/tex]:

[tex]\[ \text{Directrix} = y = -4 - 2 = -6 \][/tex]

### Summary
Thus, the focus of the parabola is at [tex]\((1, -2)\)[/tex] and the directrix is the line [tex]\(y = -6\)[/tex].

[tex]\[ \text{Focus: } (1, -2) \][/tex]
[tex]\[ \text{Directrix: } y = -6 \][/tex]
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