Eliminate the Parameter
-$x = 3 \tan t , y = 4 \sec t ; 0 \leq t \leq 2 \pi$
A) $\frac { y ^ { 2 } } { 16 } - \frac { x ^ { 2 } } { 9 } = 1 ; - \infty

Question

Eliminate the Parameter
-$x = 3 \tan t , y = 4 \sec t ; 0 \leq t \leq 2 \pi$
A) $\frac { y ^ { 2 } } { 16 } - \frac { x ^ { 2 } } { 9 } = 1 ; - \infty < x < \infty$
B) $\frac { y ^ { 2 } } { 16 } + \frac { x ^ { 2 } } { 9 } = 1 ; - \infty < x < \infty$
C) $y = 4 \sqrt { 1 + \frac { x ^ { 2 } } { 9 } } ; - \infty < x < \infty$
D) $y = x ^ { 2 } - 9 ; - 3 \leq x \leq 3$

Quizplus · Accepted Answer

Using the trigonometric identity $\sec^2 t = 1 + \tan^2 t$, we can eliminate the parameter $t$ by expressing both $x$ and $y$ in terms of trigonometric functions of $t$. Given $x = 3\tan t$ and $y = 4\sec t$, we can write $\tan t = \frac{x}{3}$ and $\sec t = \frac{y}{4}$. Squaring both equations and using the identity, we get $\frac{x^2}{9} + 1 = \frac{y^2}{16}$, which simplifies to $\frac{y^2}{16} - \frac{x^2}{9} = 1$, matching choice A.

Eliminate the Parameter - x=3tan⁡t,y=4sec⁡t;0≤t≤2πx = 3 \tan t , y = 4 \sec t ; 0 \leq t \leq 2 \pix=3tant,y=4sect;0≤t≤2π

Eliminate the Parameter
- $x = 3 \tan t , y = 4 \sec t ; 0 \leq t \leq 2 \pi$