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To find the equation of the graph that represents all points in the plane that are equidistant from the point [tex]\( F(0, -8) \)[/tex] and the line [tex]\( y = 8 \)[/tex], we use the definition of a parabola. For any point [tex]\((x, y)\)[/tex] on the parabola, the distance to the focus [tex]\( F(0, -8) \)[/tex] should be equal to its distance to the directrix [tex]\( y = 8 \)[/tex].
### Step-by-Step Solution:
1. Express the distance from [tex]\((x, y)\)[/tex] to the focus [tex]\( F(0, -8) \)[/tex]:
The distance formula between [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:
[tex]\[ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
So, the distance from [tex]\((x, y)\)[/tex] to [tex]\( F(0, -8) \)[/tex] is:
[tex]\[ d_1 = \sqrt{(x - 0)^2 + (y - (-8))^2} = \sqrt{x^2 + (y + 8)^2} \][/tex]
2. Express the distance from [tex]\((x, y)\)[/tex] to the directrix [tex]\( y = 8 \)[/tex]:
The distance from a point [tex]\((x, y)\)[/tex] to the line [tex]\( y = k \)[/tex] is given by the vertical distance formula:
[tex]\[ d_2 = |y - k| \][/tex]
Therefore, the distance from [tex]\((x, y)\)[/tex] to the line [tex]\( y = 8 \)[/tex] is:
[tex]\[ d_2 = |y - 8| \][/tex]
3. Set the two distances equal to each other:
According to the definition of a parabola:
[tex]\[ d_1 = d_2 \][/tex]
Thus, we have:
[tex]\[ \sqrt{x^2 + (y + 8)^2} = |y - 8| \][/tex]
4. Remove the absolute value and square both sides to eliminate the square root:
To handle the absolute value, we consider two cases for simplification. Squaring both sides will help remove the square root:
[tex]\[ x^2 + (y + 8)^2 = (y - 8)^2 \][/tex]
5. Expand both sides:
[tex]\[ x^2 + y^2 + 16y + 64 = y^2 - 16y + 64 \][/tex]
6. Simplify the equation:
Cancel [tex]\( y^2 \)[/tex] and [tex]\( 64 \)[/tex] from both sides:
[tex]\[ x^2 + 16y = -16y \][/tex]
7. Combine like terms:
[tex]\[ x^2 + 32y = 0 \][/tex]
8. Solve for [tex]\( y \)[/tex]:
[tex]\[ y = -\frac{1}{32} x^2 \][/tex]
Thus, the equation for the parabola where each point [tex]\((x, y)\)[/tex] is equidistant from the point [tex]\( F(0, -8) \)[/tex] and the line [tex]\( y = 8 \)[/tex] is:
[tex]\[ y = -\frac{1}{32} x^2 \][/tex]
There is no need for the equation
[tex]\[ -\frac{1}{8} x^2 \][/tex]
So the final answer is:
[tex]\[ y = -\frac{1}{32} x^2 \][/tex]
### Step-by-Step Solution:
1. Express the distance from [tex]\((x, y)\)[/tex] to the focus [tex]\( F(0, -8) \)[/tex]:
The distance formula between [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:
[tex]\[ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
So, the distance from [tex]\((x, y)\)[/tex] to [tex]\( F(0, -8) \)[/tex] is:
[tex]\[ d_1 = \sqrt{(x - 0)^2 + (y - (-8))^2} = \sqrt{x^2 + (y + 8)^2} \][/tex]
2. Express the distance from [tex]\((x, y)\)[/tex] to the directrix [tex]\( y = 8 \)[/tex]:
The distance from a point [tex]\((x, y)\)[/tex] to the line [tex]\( y = k \)[/tex] is given by the vertical distance formula:
[tex]\[ d_2 = |y - k| \][/tex]
Therefore, the distance from [tex]\((x, y)\)[/tex] to the line [tex]\( y = 8 \)[/tex] is:
[tex]\[ d_2 = |y - 8| \][/tex]
3. Set the two distances equal to each other:
According to the definition of a parabola:
[tex]\[ d_1 = d_2 \][/tex]
Thus, we have:
[tex]\[ \sqrt{x^2 + (y + 8)^2} = |y - 8| \][/tex]
4. Remove the absolute value and square both sides to eliminate the square root:
To handle the absolute value, we consider two cases for simplification. Squaring both sides will help remove the square root:
[tex]\[ x^2 + (y + 8)^2 = (y - 8)^2 \][/tex]
5. Expand both sides:
[tex]\[ x^2 + y^2 + 16y + 64 = y^2 - 16y + 64 \][/tex]
6. Simplify the equation:
Cancel [tex]\( y^2 \)[/tex] and [tex]\( 64 \)[/tex] from both sides:
[tex]\[ x^2 + 16y = -16y \][/tex]
7. Combine like terms:
[tex]\[ x^2 + 32y = 0 \][/tex]
8. Solve for [tex]\( y \)[/tex]:
[tex]\[ y = -\frac{1}{32} x^2 \][/tex]
Thus, the equation for the parabola where each point [tex]\((x, y)\)[/tex] is equidistant from the point [tex]\( F(0, -8) \)[/tex] and the line [tex]\( y = 8 \)[/tex] is:
[tex]\[ y = -\frac{1}{32} x^2 \][/tex]
There is no need for the equation
[tex]\[ -\frac{1}{8} x^2 \][/tex]
So the final answer is:
[tex]\[ y = -\frac{1}{32} x^2 \][/tex]
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