To find the foci of the hyperbola represented by the equation [tex]\(5y^2 - 4x^2 - 80 = 0\)[/tex], follow these detailed steps:
1. Rewrite the equation in standard form:
We start with the given equation:
[tex]\[ 5y^2 - 4x^2 - 80 = 0 \][/tex]
Add 80 to both sides to isolate the quadratic terms:
[tex]\[ 5y^2 - 4x^2 = 80 \][/tex]
2. Divide by 80 to normalize the equation:
[tex]\[ \frac{5y^2}{80} - \frac{4x^2}{80} = \frac{80}{80} \][/tex]
Simplify the fractions:
[tex]\[ \frac{y^2}{16} - \frac{x^2}{20} = 1 \][/tex]
3. Identify the values of [tex]\(a^2\)[/tex] and [tex]\(b^2\)[/tex]:
The equation in standard form for a hyperbola [tex]\(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\)[/tex] gives us:
[tex]\[ a^2 = 16 \][/tex]
[tex]\[ b^2 = 20 \][/tex]
4. Calculate [tex]\(c\)[/tex] (the distance from the center to the foci):
For hyperbolas of the form [tex]\(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\)[/tex], the relationship between [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] is given by:
[tex]\[ c^2 = a^2 + b^2 \][/tex]
Substitute the values:
[tex]\[ c^2 = 16 + 20 \][/tex]
[tex]\[ c^2 = 36 \][/tex]
[tex]\[ c = \sqrt{36} = 6 \][/tex]
5. Determine the coordinates of the foci:
As the hyperbola opens vertically (since the [tex]\(y^2\)[/tex] term comes first), the foci will be located along the [tex]\(y\)[/tex]-axis:
[tex]\[ (0, \pm c) \][/tex]
Using the value we found for [tex]\(c\)[/tex]:
[tex]\[ (0, 6) \][/tex]
[tex]\[ (0, -6) \][/tex]
So, the foci of the hyperbola represented by the equation [tex]\(5y^2 - 4x^2 - 80 = 0\)[/tex] are [tex]\((0, 6)\)[/tex] and [tex]\((0, -6)\)[/tex].
The correct answer is:
[tex]\((0,-6)\)[/tex] and [tex]\((0,6)\)[/tex].
