Question

1. (20 points) find the center, vertices, foci, and asymptotes of the hyperbola with equation:

\\81x^2 - 16y^2 + 324x - 128y - 1228 = 0\\

and graph the hyperbola.

center:
vertices:
foci:
asymptotes:

Explanation:

Step1: Group terms

Group the \(x\) and \(y\) terms:

$$(81x^2 + 324x) - (16y^2 + 128y) = 1228$$

Step2: Factor coefficients

Factor out the leading coefficients:

$$81(x^2 + 4x) - 16(y^2 + 8y) = 1228$$

Step3: Complete squares

Add necessary constants inside parentheses:

$$81(x^2 + 4x + 4) - 16(y^2 + 8y + 16) = 1228 + 324 - 256$$

Step4: Simplify equation

Simplify the right side:

$$81(x + 2)^2 - 16(y + 4)^2 = 1296$$

Step5: Standard form

Divide by 1296:

$$\frac{(x + 2)^2}{16} - \frac{(y + 4)^2}{81} = 1$$

Step6: Identify parameters

Find \(h\), \(k\), \(a\), \(b\), and \(c\):

$$h = -2, \quad k = -4, \quad a = 4, \quad b = 9, \quad c = \sqrt{16 + 81} = \sqrt{97}$$

Step7: Determine properties

Calculate center, vertices, foci, and asymptotes:

$$\text{Center: } (-2, -4)$$

$$\text{Vertices: } (-2 \pm 4, -4) \Rightarrow (-6, -4), (2, -4)$$

$$\text{Foci: } (-2 \pm \sqrt{97}, -4)$$

$$\text{Asymptotes: } y + 4 = \pm \frac{9}{4}(x + 2)$$

Answer:

Center: \((-2, -4)\)
Vertices: \((-6, -4)\) and \((2, -4)\)
Foci: \((-2 - \sqrt{97}, -4)\) and \((-2 + \sqrt{97}, -4)\)
Asymptotes: \(y + 4 = \pm \frac{9}{4}(x + 2)\)