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Sagot :
To solve the problem of finding the center, vertices, foci, and asymptotes for the given equation of the hyperbola [tex]\(4y^2 - x^2 = 36\)[/tex], let's follow these steps:
1. Rewrite the equation in standard form:
[tex]\[ 4y^2 - x^2 = 36 \][/tex]
Divide both sides by 36 to get it into the standard form of a hyperbola:
[tex]\[ \frac{4y^2}{36} - \frac{x^2}{36} = \frac{36}{36} \][/tex]
Simplify the equation:
[tex]\[ \frac{y^2}{9} - \frac{x^2}{36} = 1 \][/tex]
2. Identify the parameters:
In the standard form of a hyperbola, [tex]\(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\)[/tex], we can identify:
[tex]\[ a^2 = 9 \quad \text{and} \quad b^2 = 36 \][/tex]
Thus,
[tex]\[ a = \sqrt{9} = 3 \quad \text{and} \quad b = \sqrt{36} = 6 \][/tex]
3. Determine the center:
The center of the hyperbola is at the origin [tex]\( (0,0) \)[/tex].
4. Find the vertices:
The vertices of a hyperbola in the form [tex]\(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\)[/tex] are located at [tex]\((0, \pm a)\)[/tex]. So, the vertices are:
[tex]\[ (0, 3) \quad \text{and} \quad (0, -3) \][/tex]
5. Calculate the foci:
The foci are located at [tex]\( (0, \pm c) \)[/tex], where [tex]\( c \)[/tex] is found using the relationship:
[tex]\[ c^2 = a^2 + b^2 \][/tex]
Substituting the known values:
[tex]\[ c^2 = 9 + 36 = 45 \][/tex]
Therefore,
[tex]\[ c = \sqrt{45} = 3\sqrt{5} \approx 6.708 \][/tex]
So, the foci are:
[tex]\[ (0, 6.708) \quad \text{and} \quad (0, -6.708) \][/tex]
6. Determine the equations of the asymptotes:
The asymptotes for a hyperbola in the form [tex]\(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\)[/tex] are given by:
[tex]\[ y = \pm \frac{a}{b} x \][/tex]
Substituting the values for [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ y = \pm \frac{3}{6} x \][/tex]
Simplify the slope:
[tex]\[ y = \pm \frac{1}{2} x \][/tex]
Therefore, the asymptotes are:
[tex]\[ y = \frac{1}{2} x \quad \text{and} \quad y = -\frac{1}{2} x \][/tex]
In summary, the properties of the hyperbola [tex]\(4y^2 - x^2 = 36\)[/tex] are:
- Center: [tex]\((0,0)\)[/tex]
- Vertices: [tex]\((0, 3)\)[/tex] and [tex]\((0, -3)\)[/tex]
- Foci: [tex]\((0, 6.708)\)[/tex] and [tex]\((0, -6.708)\)[/tex]
- Asymptotes: [tex]\(y = \frac{1}{2} x\)[/tex] and [tex]\(y = -\frac{1}{2} x\)[/tex]
1. Rewrite the equation in standard form:
[tex]\[ 4y^2 - x^2 = 36 \][/tex]
Divide both sides by 36 to get it into the standard form of a hyperbola:
[tex]\[ \frac{4y^2}{36} - \frac{x^2}{36} = \frac{36}{36} \][/tex]
Simplify the equation:
[tex]\[ \frac{y^2}{9} - \frac{x^2}{36} = 1 \][/tex]
2. Identify the parameters:
In the standard form of a hyperbola, [tex]\(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\)[/tex], we can identify:
[tex]\[ a^2 = 9 \quad \text{and} \quad b^2 = 36 \][/tex]
Thus,
[tex]\[ a = \sqrt{9} = 3 \quad \text{and} \quad b = \sqrt{36} = 6 \][/tex]
3. Determine the center:
The center of the hyperbola is at the origin [tex]\( (0,0) \)[/tex].
4. Find the vertices:
The vertices of a hyperbola in the form [tex]\(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\)[/tex] are located at [tex]\((0, \pm a)\)[/tex]. So, the vertices are:
[tex]\[ (0, 3) \quad \text{and} \quad (0, -3) \][/tex]
5. Calculate the foci:
The foci are located at [tex]\( (0, \pm c) \)[/tex], where [tex]\( c \)[/tex] is found using the relationship:
[tex]\[ c^2 = a^2 + b^2 \][/tex]
Substituting the known values:
[tex]\[ c^2 = 9 + 36 = 45 \][/tex]
Therefore,
[tex]\[ c = \sqrt{45} = 3\sqrt{5} \approx 6.708 \][/tex]
So, the foci are:
[tex]\[ (0, 6.708) \quad \text{and} \quad (0, -6.708) \][/tex]
6. Determine the equations of the asymptotes:
The asymptotes for a hyperbola in the form [tex]\(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\)[/tex] are given by:
[tex]\[ y = \pm \frac{a}{b} x \][/tex]
Substituting the values for [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ y = \pm \frac{3}{6} x \][/tex]
Simplify the slope:
[tex]\[ y = \pm \frac{1}{2} x \][/tex]
Therefore, the asymptotes are:
[tex]\[ y = \frac{1}{2} x \quad \text{and} \quad y = -\frac{1}{2} x \][/tex]
In summary, the properties of the hyperbola [tex]\(4y^2 - x^2 = 36\)[/tex] are:
- Center: [tex]\((0,0)\)[/tex]
- Vertices: [tex]\((0, 3)\)[/tex] and [tex]\((0, -3)\)[/tex]
- Foci: [tex]\((0, 6.708)\)[/tex] and [tex]\((0, -6.708)\)[/tex]
- Asymptotes: [tex]\(y = \frac{1}{2} x\)[/tex] and [tex]\(y = -\frac{1}{2} x\)[/tex]
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