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Find the focus and directrix of the following parabola:
[tex]\[
(y-1)^2=4(x-1)
\][/tex]

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

Directrix: [tex]\(x = \square\)[/tex]

Sagot :

To find the focus and directrix of the given parabola equation:

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

we should start by rewriting it in the standard form of a parabola that opens horizontally. The standard form is:

[tex]\[ (y - k)^2 = 4p(x - h) \][/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 (and also to the directrix).

In this equation:
- Matching [tex]\((y - 1)^2 = 4(x - 1)\)[/tex] to the standard form, we identify [tex]\(h = 1\)[/tex], [tex]\(k = 1\)[/tex], and [tex]\(4p = 4\)[/tex].

We solve for [tex]\(p\)[/tex]:
- [tex]\(p = \frac{4}{4} = 1\)[/tex]

The focus of the parabola is found using [tex]\((h + p, k)\)[/tex]:
- [tex]\(h + p = 1 + 1 = 2\)[/tex]
- [tex]\(k = 1\)[/tex]

So, the coordinates of the focus are:
[tex]\[ (2, 1) \][/tex]

The directrix is a vertical line given by [tex]\(x = h - p\)[/tex]:
- [tex]\(h - p = 1 - 1 = 0\)[/tex]

So, the directrix is:
[tex]\[ x = 0 \][/tex]

Therefore, the focus and directrix of the parabola [tex]\((y - 1)^2 = 4(x - 1)\)[/tex] are:

Focus: [tex]\((2, 1)\)[/tex]

Directrix: [tex]\(x = 0\)[/tex]

So the final answers are:

[tex]\[ \text{Focus: } (2, 1) \][/tex]
[tex]\[ \text{Directrix: } x = 0 \][/tex]