Eliminate the Parameter
-A hyperbola: $x = 1 + 4 \sec t , y = 5 + 5$ tan $t$
A) $\frac { ( x - 1 ) ^ { 2 } } { 16 } - \frac { ( y - 5 ) ^ { 2 } } { 25 } = 1$
B) $\frac { ( x - 1 ) ^ { 2 } } { 16 } + \frac { ( y - 5 ) ^ { 2 } } { 25 } = 1$
C) $\frac { ( x + 5 ) ^ { 2 } } { 16 } - \frac { ( y + 1 ) ^ { 2 } } { 25 } = 1$
D) $\frac { ( x - 5 ) ^ { 2 } } { 25 } + \frac { ( y - 1 ) ^ { 2 } } { 16 } = 1$

Question

Quizplus · Accepted Answer

Using the trigonometric identity $\sec^2 t - \tan^2 t = 1$, we can eliminate the parameter $t$ from the given equations. Substituting $x = 1 + 4\sec t$ and $y = 5 + 5\tan t$ into the identity gives $\frac{(x-1)^2}{16} - \frac{(y-5)^2}{25} = 1$, which matches option A.

Eliminate the Parameter -A Hyperbola x=1+4sec⁡t,y=5+5x = 1 + 4 \sec t , y = 5 + 5x=1+4sect,y=5+5

Eliminate the Parameter
-A Hyperbola $x = 1 + 4 \sec t , y = 5 + 5$