Question

Which of the following two sets of parametric functions both represent the same ellipse. Explain the difference between the graphs.
x=3 cos t and y=8 sin t
X=3 cos 4t and y = 8 sin 4t

Answer 1

the answer:
the main equation parametric of an  ellipse is 
x²/a² + y²/b² = 1
a≠0 and b≠0

let's consider x=3 cos t and y=8 sin t, these are equivalent to x²=9 cos²t and y²=64 sin²t, and imiplying  x²/9=cos²t and y²/64=sin²t
therefore, x²/9+y²/64= cos²t+ sin²t, but we know that cos²t+ sin²t =1 (trigonometric fundamental rule)
so finally,  x²/9+y²/64=1 equivalent of  x²/3²+y²/8²= 1
this is an ellipse
with the same method, we found
x²/9+y²/64= cos²4t+ sin²4t =1, so the only difference between the graphs is the value of the angle (t and 4t)

Answer 2

Answer with explanation:

The parametric equation of same Ellipse is

  x=3 cos t and y=8 sin t

x=3 cos 4 t and y = 8 sin 4 t

The equation of the ellipse is

  $\rightarrow[\frac{x}{3}]^2+[\frac{y}{8}]^2=\cos^2 t +\sin^2 t \text{or} \cos^2 4t +\sin^2 4t\\\\\frac{x^2}{9}+\frac{y^2}{64}=1$

The difference here is

 $\frac{y}{x}=\frac{8 \tan t}{3}\\\\y=\frac{8\tan t\times x}{3}\\\\\frac{y}{x}=\frac{8 \tan 4t}{3}\\\\y=\frac{8\tan 4t\times x}{3}$

Both the parametric equation represent lines in two variable, both passing through origin.

→Slope of , parametric function, having period , (- π, π), that is x=3 cos t and y=8 sin t

$\frac{8\tan t}{3}$

→And slope of parametric function,x=3 cos 4t and y = 8 sin 4t having period

   $(\frac{-\pi}{4}, \frac{\pi}{4}) \text{is} \frac{8\tan 4t}{3}.$