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Sagot :
To find the focus of the parabola given by the equation [tex]\( y=-\frac{1}{4}x^2-2x-2 \)[/tex], we need to follow these steps:
1. Rewrite the equation in vertex form: The standard form of a parabola is [tex]\( y = a(x - h)^2 + k \)[/tex], where [tex]\((h, k)\)[/tex] is the vertex of the parabola.
Start with the given equation:
[tex]\[ y = -\frac{1}{4}x^2 - 2x - 2 \][/tex]
First, factor out [tex]\(-\frac{1}{4}\)[/tex] from the terms involving [tex]\(x\)[/tex]:
[tex]\[ y = -\frac{1}{4}(x^2 + 8x) - 2 \][/tex]
To complete the square inside the parentheses, we add and subtract the square of half the coefficient of [tex]\(x\)[/tex], which is [tex]\(4\)[/tex]:
[tex]\[ x^2 + 8x = (x + 4)^2 - 16 \][/tex]
Substitute this back into the equation:
[tex]\[ y = -\frac{1}{4}((x + 4)^2 - 16) - 2 \][/tex]
Simplify by distributing [tex]\(-\frac{1}{4}\)[/tex]:
[tex]\[ y = -\frac{1}{4}(x + 4)^2 + 4 - 2 \][/tex]
[tex]\[ y = -\frac{1}{4}(x + 4)^2 + 2 \][/tex]
Therefore, the vertex form of the equation is:
[tex]\[ y = -\frac{1}{4}(x + 4)^2 + 2 \][/tex]
Hence, the vertex [tex]\((h, k)\)[/tex] of the parabola is [tex]\((-4, 2)\)[/tex].
2. Determine the focus: The focus of a parabola [tex]\(y = a(x - h)^2 + k\)[/tex] can be found using the parameter [tex]\(a\)[/tex].
The distance from the vertex to the focus (and also from the vertex to the directrix) is given by:
[tex]\[ p = \frac{1}{4a} \][/tex]
For our equation, [tex]\( a = -\frac{1}{4} \)[/tex]:
[tex]\[ p = \frac{1}{4 \left(-\frac{1}{4}\right)} = \frac{1}{-1} = -1 \][/tex]
Since the parabola opens downward (because [tex]\( a < 0 \)[/tex]), the focus will be below the vertex. The [tex]\(y\)[/tex]-coordinate of the focus is:
[tex]\[ y = k + p = 2 - 1 = 1 \][/tex]
Therefore, the focus of the parabola is:
[tex]\((-4, 1)\)[/tex].
Hence, the coordinates of the focus are:
[tex]\[ \boxed{-4, 1.0} \][/tex]
1. Rewrite the equation in vertex form: The standard form of a parabola is [tex]\( y = a(x - h)^2 + k \)[/tex], where [tex]\((h, k)\)[/tex] is the vertex of the parabola.
Start with the given equation:
[tex]\[ y = -\frac{1}{4}x^2 - 2x - 2 \][/tex]
First, factor out [tex]\(-\frac{1}{4}\)[/tex] from the terms involving [tex]\(x\)[/tex]:
[tex]\[ y = -\frac{1}{4}(x^2 + 8x) - 2 \][/tex]
To complete the square inside the parentheses, we add and subtract the square of half the coefficient of [tex]\(x\)[/tex], which is [tex]\(4\)[/tex]:
[tex]\[ x^2 + 8x = (x + 4)^2 - 16 \][/tex]
Substitute this back into the equation:
[tex]\[ y = -\frac{1}{4}((x + 4)^2 - 16) - 2 \][/tex]
Simplify by distributing [tex]\(-\frac{1}{4}\)[/tex]:
[tex]\[ y = -\frac{1}{4}(x + 4)^2 + 4 - 2 \][/tex]
[tex]\[ y = -\frac{1}{4}(x + 4)^2 + 2 \][/tex]
Therefore, the vertex form of the equation is:
[tex]\[ y = -\frac{1}{4}(x + 4)^2 + 2 \][/tex]
Hence, the vertex [tex]\((h, k)\)[/tex] of the parabola is [tex]\((-4, 2)\)[/tex].
2. Determine the focus: The focus of a parabola [tex]\(y = a(x - h)^2 + k\)[/tex] can be found using the parameter [tex]\(a\)[/tex].
The distance from the vertex to the focus (and also from the vertex to the directrix) is given by:
[tex]\[ p = \frac{1}{4a} \][/tex]
For our equation, [tex]\( a = -\frac{1}{4} \)[/tex]:
[tex]\[ p = \frac{1}{4 \left(-\frac{1}{4}\right)} = \frac{1}{-1} = -1 \][/tex]
Since the parabola opens downward (because [tex]\( a < 0 \)[/tex]), the focus will be below the vertex. The [tex]\(y\)[/tex]-coordinate of the focus is:
[tex]\[ y = k + p = 2 - 1 = 1 \][/tex]
Therefore, the focus of the parabola is:
[tex]\((-4, 1)\)[/tex].
Hence, the coordinates of the focus are:
[tex]\[ \boxed{-4, 1.0} \][/tex]
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