Find the focus and directrix of the following parabola:

[tex](y-4)^2 = 8(x-3)[/tex]

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

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



Answer :

To find the focus and directrix of the parabola given by the equation [tex]\((y-4)^2 = 8(x-3)\)[/tex], let's follow a step-by-step process:

1. Identify the standard form of the parabola:
The given equation is of the form [tex]\((y - k)^2 = 4p(x - h)\)[/tex], which represents a parabola that opens horizontally.

2. Match the given equation to the standard form:
Comparing [tex]\((y - 4)^2 = 8(x - 3)\)[/tex] with [tex]\((y - k)^2 = 4p(x - h)\)[/tex], we can identify the following parameters:
- [tex]\( h = 3 \)[/tex]
- [tex]\( k = 4 \)[/tex]
- [tex]\( 4p = 8 \)[/tex]

3. Solve for [tex]\( p \)[/tex]:
From the equation [tex]\(4p = 8\)[/tex], we solve for [tex]\( p \)[/tex]:
[tex]\[ p = \frac{8}{4} = 2 \][/tex]

4. Find the coordinates of the focus:
The focus of a parabola in the form [tex]\((y - k)^2 = 4p(x - h)\)[/tex] is located at [tex]\((h + p, k)\)[/tex].
Given [tex]\( h = 3 \)[/tex] and [tex]\( k = 4 \)[/tex], and having determined [tex]\( p = 2 \)[/tex], we find:
[tex]\[ \text{Focus} = (h + p, k) = (3 + 2, 4) = (5, 4) \][/tex]

5. Find the equation of the directrix:
The directrix of a parabola in the form [tex]\((y - k)^2 = 4p(x - h)\)[/tex] is the vertical line given by [tex]\( x = h - p \)[/tex].
Given [tex]\( h = 3 \)[/tex] and [tex]\( p = 2 \)[/tex], we find:
[tex]\[ \text{Directrix} = x = h - p = 3 - 2 = 1 \][/tex]

Therefore, the focus of the parabola [tex]\((y-4)^2 = 8(x-3)\)[/tex] is at [tex]\((5, 4)\)[/tex], and the directrix is the line [tex]\(x = 1\)[/tex].

Final results:

- Focus: [tex]\((5, 4)\)[/tex]
- Directrix: [tex]\(x = 1\)[/tex]