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Which rectangular equation represents the parametric equations [tex]x=2 \sec (t)+1[/tex] and [tex]y=6 \tan (t)[/tex]?

A. [tex]\frac{(x-1)^2}{4}-\frac{y^2}{36}=1[/tex]
B. [tex]\frac{(x-1)^2}{4}+\frac{y^2}{36}=1[/tex]
C. [tex]\frac{(x-1)^2}{4}-\frac{y^2}{6}=1[/tex]
D. [tex]\frac{(x-1)^2}{4}+\frac{y^2}{6}=1[/tex]


Sagot :

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]