To find the rectangular equation representing the given parametric equations [tex]\( x = 2 \sec(t) + 1 \)[/tex] and [tex]\( y = 6 \tan(t) \)[/tex], we need to eliminate the parameter [tex]\( t \)[/tex].
1. Express [tex]\(\sec(t)\)[/tex] and [tex]\(\tan(t)\)[/tex] in terms of [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
- For [tex]\(x = 2 \sec(t) + 1\)[/tex]:
[tex]\[
x - 1 = 2 \sec(t)
\][/tex]
[tex]\[
\sec(t) = \frac{x - 1}{2}
\][/tex]
- For [tex]\(y = 6 \tan(t)\)[/tex]:
[tex]\[
\tan(t) = \frac{y}{6}
\][/tex]
2. Use the Pythagorean identity [tex]\(\sec^2(t) - \tan^2(t) = 1\)[/tex] to eliminate [tex]\(t\)[/tex]:
[tex]\[
\sec^2(t) - \tan^2(t) = 1
\][/tex]
Substitute [tex]\(\sec(t)\)[/tex] and [tex]\(\tan(t)\)[/tex] from above:
[tex]\[
\left(\frac{x - 1}{2}\right)^2 - \left(\frac{y}{6}\right)^2 = 1
\][/tex]
3. Simplify the equation:
[tex]\[
\left(\frac{x - 1}{2}\right)^2 - \left(\frac{y}{6}\right)^2 = 1
\][/tex]
Square the fractions:
[tex]\[
\frac{(x - 1)^2}{4} - \frac{y^2}{36} = 1
\][/tex]
This is the equation of a hyperbola.
Therefore, the correct rectangular equation representing the given parametric equations is:
[tex]\[
\boxed{\frac{(x-1)^2}{4}-\frac{y^2}{36}=1}
\][/tex]
