Parametric form of ellipse $x^2 + 16y^2 = 4$

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 $$x=2\cos \theta\quad y=\frac{1}{2}\sin \theta $$