Question 185921
*[Tex \LARGE x=2\sec(t)] and *[Tex \LARGE y=4\tan(t)] ... Start with the given parametric equations.



*[Tex \LARGE x^2=(2\sec(t))^2] and *[Tex \LARGE y^2=(4\tan(t))^2] ... Square both sides for each equation.




*[Tex \LARGE x^2=4\sec^2(t)] and *[Tex \LARGE y^2=16\tan^2(t)] ... Square the right sides and simplify




*[Tex \LARGE \frac{x^2}{4}=\sec^2(t)] and *[Tex \LARGE \frac{y^2}{16}=\tan^2(t)] ... Isolate for *[Tex \LARGE \sec^2(t)] and *[Tex \LARGE \tan^2(t)].



*[Tex \LARGE \sec^2(t)=\frac{x^2}{4}] and *[Tex \LARGE \tan^2(t)=\frac{y^2}{16}] Rearrange the equations




Now remember, *[Tex \LARGE 1+\tan^2(t)=\sec^2(t)] is an identity



*[Tex \LARGE 1+\tan^2(t)=\sec^2(t)] ... Start with the given identity



*[Tex \LARGE 1+\frac{y^2}{16}=\frac{x^2}{4}] ... Plug in *[Tex \LARGE \sec^2(t)=\frac{x^2}{4}] and *[Tex \LARGE \tan^2(t)=\frac{y^2}{16}]



*[Tex \LARGE 1=\frac{x^2}{4}-\frac{y^2}{16}] ... Subtract *[Tex \LARGE \frac{y^2}{16}] from both sides.



*[Tex \LARGE \frac{x^2}{4}-\frac{y^2}{16}=1] Rearrange the equation.



Now the equation is in rectangular form 



Note: the last equation is in the form of a hyperbola.