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Find the center, the vertices, the foci, and the asymptotes of the hyperbola.

[tex]\[ 4y^2 - x^2 = 36 \][/tex]

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]