The vertices are (±1/2, 0), the asymptote is y = -6x, and the foci of the hyperbola is [tex](\±\frac{\sqrt{37}}{2}, 0)[/tex]
How to determine the vertices, asymptotes, and the foci?
The equation of the hyperbola is given as:
576x^2 - 16y^2 = 144
Divide through by 144
[tex]4x^2 - \frac{y^2}{9} = 1[/tex]
Rewrite as:
[tex]\frac{x^2}{1/4} - \frac{y^2}{9} = 1[/tex]
Express as squares
[tex]\frac{x^2}{1/2^2} - \frac{y^2}{3^2} = 1[/tex]
The equation of a hyperbola is:
[tex]\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1[/tex]
So, we have:
a = 1/2 and b = 3
Calculate c using:
c = √(a² + b²)
So, we have:
c = √(1/4 + 9)
Evaluate the sum
[tex]c = \sqrt{\frac{37}{4}}[/tex]
[tex]c = \frac{\sqrt{37}}2[/tex]
The coordinates of the vertices is:
Vertices = (±a, 0)
So, we have:
Vertices = (±1/2, 0)
The coordinates of the foci are:
Foci = (±c, 0)
So, we have:
[tex]Foci = (\±\frac{\sqrt{37}}{2}, 0)[/tex]
The asymptote of the hyperbola is:
[tex]y =- \frac{b}{a} *x[/tex]
This gives
[tex]y = -\frac{3}{1/2} *x[/tex]
Evaluate
y = -6x
Hence, the vertices are (±1/2, 0), the asymptote is y = -6x, and the foci of the hyperbola is [tex](\±\frac{\sqrt{37}}{2}, 0)[/tex]
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