Answer :
To find the equations representing the asymptotes of the hyperbola [tex]\(\frac{(x-1)^2}{36} - \frac{(y-2)^2}{64} = 1\)[/tex], we should use the following process:
1. Identify the general form of the hyperbola equation:
[tex]\[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \][/tex]
Here, [tex]\((h, k)\)[/tex] is the center of the hyperbola, [tex]\(a^2\)[/tex] and [tex]\(b^2\)[/tex] are the denominators of the respective terms.
2. Determine the parameters:
- The center [tex]\((h, k)\)[/tex] is [tex]\((1, 2)\)[/tex].
- [tex]\(a^2 = 36 \Rightarrow a = 6\)[/tex].
- [tex]\(b^2 = 64 \Rightarrow b = 8\)[/tex].
3. Write the equations for the asymptotes:
The asymptotes of a hyperbola in this form are given by:
[tex]\[ y - k = \pm \frac{b}{a}(x - h) \][/tex]
Substitute [tex]\(b = 8\)[/tex], [tex]\(a = 6\)[/tex], [tex]\(h = 1\)[/tex], and [tex]\(k = 2\)[/tex] into the asymptote equations.
For the positive slope:
[tex]\[ y - 2 = \frac{8}{6}(x - 1) \][/tex]
Simplify it:
[tex]\[ y - 2 = \frac{4}{3}(x - 1) \][/tex]
[tex]\[ y - 2 = \frac{4}{3}x - \frac{4}{3} \][/tex]
[tex]\[ y = \frac{4}{3}x - \frac{4}{3} + 2 \][/tex]
[tex]\[ y = \frac{4}{3}x - \frac{4}{3} + \frac{6}{3} \][/tex]
[tex]\[ y = \frac{4}{3}x + \frac{2}{3} \][/tex]
For the negative slope:
[tex]\[ y - 2 = -\frac{8}{6}(x - 1) \][/tex]
Simplify it:
[tex]\[ y - 2 = -\frac{4}{3}(x - 1) \][/tex]
[tex]\[ y - 2 = -\frac{4}{3}x + \frac{4}{3} \][/tex]
[tex]\[ y = -\frac{4}{3}x + \frac{4}{3} + 2 \][/tex]
[tex]\[ y = -\frac{4}{3}x + \frac{4}{3} + \frac{6}{3} \][/tex]
[tex]\[ y = -\frac{4}{3}x + \frac{10}{3} \][/tex]
4. Match these equations to the given options:
The correct forms of the asymptote equations are:
[tex]\[ y = \frac{4}{3}x + \frac{2}{3} \][/tex]
[tex]\[ y = -\frac{4}{3}x + \frac{10}{3} \][/tex]
Review the provided options:
A. [tex]\(y = \frac{4x + 2}{3}\)[/tex] and [tex]\(y = \frac{10 - 4x}{3}\)[/tex]
B. [tex]\(y = \frac{3x + 2}{4}\)[/tex] and [tex]\(y = \frac{10 - 4x}{3}\)[/tex]
C. [tex]\(y = \frac{10 - 4x}{3}\)[/tex] and [tex]\(y = \frac{4x - 2}{3}\)[/tex]
D. [tex]\(y = \frac{8 + 3x}{4}\)[/tex] and [tex]\(y = \frac{4x + 2}{3}\)[/tex]
From these equations, the correct match is option:
A. [tex]\(y = \frac{4x + 2}{3}\)[/tex] and [tex]\(y = \frac{10 - 4x}{3}\)[/tex]
So, the correct answer is:
[tex]\[ \boxed{2} \][/tex]
1. Identify the general form of the hyperbola equation:
[tex]\[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \][/tex]
Here, [tex]\((h, k)\)[/tex] is the center of the hyperbola, [tex]\(a^2\)[/tex] and [tex]\(b^2\)[/tex] are the denominators of the respective terms.
2. Determine the parameters:
- The center [tex]\((h, k)\)[/tex] is [tex]\((1, 2)\)[/tex].
- [tex]\(a^2 = 36 \Rightarrow a = 6\)[/tex].
- [tex]\(b^2 = 64 \Rightarrow b = 8\)[/tex].
3. Write the equations for the asymptotes:
The asymptotes of a hyperbola in this form are given by:
[tex]\[ y - k = \pm \frac{b}{a}(x - h) \][/tex]
Substitute [tex]\(b = 8\)[/tex], [tex]\(a = 6\)[/tex], [tex]\(h = 1\)[/tex], and [tex]\(k = 2\)[/tex] into the asymptote equations.
For the positive slope:
[tex]\[ y - 2 = \frac{8}{6}(x - 1) \][/tex]
Simplify it:
[tex]\[ y - 2 = \frac{4}{3}(x - 1) \][/tex]
[tex]\[ y - 2 = \frac{4}{3}x - \frac{4}{3} \][/tex]
[tex]\[ y = \frac{4}{3}x - \frac{4}{3} + 2 \][/tex]
[tex]\[ y = \frac{4}{3}x - \frac{4}{3} + \frac{6}{3} \][/tex]
[tex]\[ y = \frac{4}{3}x + \frac{2}{3} \][/tex]
For the negative slope:
[tex]\[ y - 2 = -\frac{8}{6}(x - 1) \][/tex]
Simplify it:
[tex]\[ y - 2 = -\frac{4}{3}(x - 1) \][/tex]
[tex]\[ y - 2 = -\frac{4}{3}x + \frac{4}{3} \][/tex]
[tex]\[ y = -\frac{4}{3}x + \frac{4}{3} + 2 \][/tex]
[tex]\[ y = -\frac{4}{3}x + \frac{4}{3} + \frac{6}{3} \][/tex]
[tex]\[ y = -\frac{4}{3}x + \frac{10}{3} \][/tex]
4. Match these equations to the given options:
The correct forms of the asymptote equations are:
[tex]\[ y = \frac{4}{3}x + \frac{2}{3} \][/tex]
[tex]\[ y = -\frac{4}{3}x + \frac{10}{3} \][/tex]
Review the provided options:
A. [tex]\(y = \frac{4x + 2}{3}\)[/tex] and [tex]\(y = \frac{10 - 4x}{3}\)[/tex]
B. [tex]\(y = \frac{3x + 2}{4}\)[/tex] and [tex]\(y = \frac{10 - 4x}{3}\)[/tex]
C. [tex]\(y = \frac{10 - 4x}{3}\)[/tex] and [tex]\(y = \frac{4x - 2}{3}\)[/tex]
D. [tex]\(y = \frac{8 + 3x}{4}\)[/tex] and [tex]\(y = \frac{4x + 2}{3}\)[/tex]
From these equations, the correct match is option:
A. [tex]\(y = \frac{4x + 2}{3}\)[/tex] and [tex]\(y = \frac{10 - 4x}{3}\)[/tex]
So, the correct answer is:
[tex]\[ \boxed{2} \][/tex]
