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The transverse axis of a hyperbola is the [tex]$x$[/tex]-axis, and the equation of the asymptotes is [tex]$y= \pm \frac{4}{3} x$[/tex].

In the equation of the given hyperbola in standard form, [tex]$a = \square$[/tex], [tex]$b = \square$[/tex], and [tex]$c = \square$[/tex].


Sagot :

To solve this problem, we need to identify the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] for the given hyperbola whose transverse axis is aligned with the [tex]\(x\)[/tex]-axis, and whose asymptotes are given by the equations [tex]\( y = \pm \frac{4}{3}x \)[/tex].

1. Determine [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:

The equations of asymptotes for a hyperbola are given by [tex]\( y = \pm \frac{a}{b}x \)[/tex]. Here, it’s provided [tex]\( \frac{a}{b} = \frac{4}{3} \)[/tex], which means [tex]\(a = 4\)[/tex] and [tex]\(b = 3\)[/tex].

2. Determine [tex]\(c\)[/tex]:

For a hyperbola, the relationship between [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] is given by the equation [tex]\( c^2 = a^2 + b^2 \)[/tex]. Plugging in the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:

[tex]\[ c = \sqrt{a^2 + b^2} \][/tex]

Given [tex]\(a = 4\)[/tex] and [tex]\(b = 3\)[/tex]:

[tex]\[ c = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \][/tex]

Thus, the values are [tex]\(a = 4\)[/tex], [tex]\(b = 3\)[/tex], and [tex]\(c = 5\)[/tex].

In summary:

- [tex]\(a = 4\)[/tex]
- [tex]\(b = 3\)[/tex]
- [tex]\(c = 5\)[/tex]

Therefore, the equation of the given hyperbola in standard form has [tex]\(a = 4\)[/tex], [tex]\(b = 3\)[/tex], and [tex]\(c = 5\)[/tex].
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