Eliminate the Parameter -$$x = 2 + \sec t , y = 5 + 2 \tan t ; 0

Question

Eliminate the Parameter
-$$x = 2 + \sec t , y = 5 + 2 \tan t ; 0 < t < \frac { \pi } { 2 }$$
A) $( x - 2 ) ^ { 2 } - \frac { ( y - 5 ) ^ { 2 } } { 4 } = 1 ; x > 3$
B) $( x - 2 ) ^ { 2 } + \frac { ( y - 5 ) ^ { 2 } } { 4 } = 1 ; 1 \leq x \leq 3$
C) $( \mathrm { y } - 2 ) ^ { 2 } - \frac { ( \mathrm { x } - 5 ) ^ { 2 } } { 4 } = 1 ; - \infty < \mathrm { x } < \infty$
D) $( x - 2 ) ^ { 2 } - ( y - 5 ) ^ { 2 } = 4 ; - \infty < x < \infty$

Quizplus · Accepted Answer

To eliminate the parameter $t$, we use the trigonometric identities $\sec^2 t = 1 + \tan^2 t$ and express $x$ and $y$ in terms of $\sec t$ and $\tan t$, respectively. Given $x = 2 + \sec t$ and $y = 5 + 2\tan t$, we solve for $\sec t$ and $\tan t$: $\sec t = x - 2$ and $\tan t = \frac{y - 5}{2}$. Substituting these into the identity $\sec^2 t = 1 + \tan^2 t$ gives $(x - 2)^2 = 1 + \left(\frac{y - 5}{2}\right)^2$, which simplifies to $(x - 2)^2 - \frac{(y - 5)^2}{4} = 1$. The condition $0 < t < \frac{\pi}{2}$ implies $\sec t > 1$, so $x > 3$.

Eliminate the Parameter - x=2+sec⁡t,y=5+2tan⁡t;0<t<π2x = 2 + \sec t , y = 5 + 2 \tan t ; 0 < t < \frac { \pi } { 2 }x=2+sect,y=5+2tant;0<t<2π​

Eliminate the Parameter
- $x = 2 + \sec t , y = 5 + 2 \tan t ; 0 < t < \frac { \pi } { 2 }$