Parametric form of ellipse $x^2 + 16y^2 = 4$ The given ellipse is $-$
$x^2 +16y^2 = 4$
I know the standard method of changing the equation of ellipse in parametric form.
Instead of complicated calculation, can we simply put $x= 2 \cos \theta , y= \frac{1}{2} \sin \theta $ to satisfy the equation. Does this still represent the same ellipse?  How do I know if my parametric equation is representing the same ellipse as given ,without using graph?
 A: In terms of 'intuition' - generally the only thing you need to be careful of this technique is that you may not obtain the whole ellipse. But upon checking the domain of $\theta$. This particular parameterisation that you have used is okay! 
A choice of parameterisation is generally from the knowledge that it will satisfy the cartesian equation and that it is sufficiently variable.
For example if I wanted to find a parameterisation for the parabola $x^2=4ay$ I could simply say 
$$x=2t^2 \ y=\frac{t^4}{a}$$
Indeed, it satisfies the equation - however this would not 'capture' the parabola for $x<0$. We need to be aware of such things when we choose a parameterisation. 
A: In the case of an ellipse, the simplest solution is to make a change of variables, in order to get the equation of a circle.
For the equation
$$ x^2 +16y^2 = 4 $$
Just consider the change in variables:
$$X = x; \quad Y = 4y $$
To  get the new equation
$$X^2 + Y^2 = 4 $$
Therefore the equation of a circle of center $(0,0)$ and ray $R=2$.
The corresponding parametric equation is
$$X=R\cos \theta\quad Y=R\sin \theta $$
For $\theta$ varying from $0$ to $2 \pi$
Which corresponds to
$$x=2\cos \theta\quad y=\frac{1}{2}\sin \theta $$
A: Well, you can make the substitution in the equation, so you get:
$$(2\cos(\theta))^2+16(\frac 1 2 \sin(\theta))^2=4.$$
Therefore your parametrization is ok.
