Understanding the Hyperbola
The equation of the hyperbola is given as:
16x² – 4y² = 64
To analyze this hyperbola, we first need to rewrite it in standard form. Dividing the entire equation by 64 gives us:
\( \frac{x²}{4} – \frac{y²}{16} = 1 \)
Identifying the Standard Form
The standard form of a hyperbola centered at the origin is:
\( \frac{x²}{a²} – \frac{y²}{b²} = 1 \)
From the rewritten equation, we can see that:
- a² = 4 (therefore, a = 2)
- b² = 16 (therefore, b = 4)
Vertices
The vertices of a hyperbola in this form are located at \((\pm a, 0)\). Thus, the vertices for our hyperbola are:
- (2, 0)
- (-2, 0)
Foci
The foci of a hyperbola are determined using the formula:
\( c = \sqrt{a² + b²} \)
Calculating:
- a² = 4
- b² = 16
- c = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}
Therefore, the foci are located at:
- (2\sqrt{5}, 0)
- (-2\sqrt{5}, 0)
Asymptotes
The equations of the asymptotes for a hyperbola in this form are given by:
y = \pm \frac{b}{a} x
Substituting our values of a and b, we find:
y = \pm \frac{4}{2} x
This simplifies to:
y = \pm 2x
Thus, the equations of the asymptotes are:
- y = 2x
- y = -2x
Conclusion
In summary, for the hyperbola given by the equation 16x² – 4y² = 64:
- Vertices: (2, 0) and (-2, 0)
- Foci: (2\sqrt{5}, 0) and (-2\sqrt{5}, 0)
- Asymptotes: y = 2x and y = -2x