How do we find the range of the binary relation $4{{x}^{2}}+9{{y}^{2}}=36$?

Hint: Now we will rearrange the terms of the equation such that we get x in terms of y. Now we will find the domain of the obtained function. We know that if we have $\sqrt{x}$ then x must be greater than or equal to zero. Hence using this condition we can easily find the range of the function.

Complete step by step answer:
Now we want to find the range of the binary relation $4{{x}^{2}}+9{{y}^{2}}=36$
Now we will first write the equation as x in terms of y and then find the domain of the obtained function
To do so we will subtract the term $9{{y}^{2}}$ on both sides of the equation.
Hence we have,
$\Rightarrow 4{{x}^{2}}=36-9{{y}^{2}}$
Now dividing the whole equation by 4 we get,
$\Rightarrow {{x}^{2}}=\dfrac{36-9{{y}^{2}}}{4}$
Now taking square root on both sides of equation we get,
$\begin{align}
  & \Rightarrow x=\sqrt{\dfrac{36-9{{y}^{2}}}{4}} \\
 & \Rightarrow x=\dfrac{\sqrt{9\left( 4-{{y}^{2}} \right)}}{2} \\
 & \Rightarrow x=\dfrac{3\sqrt{4-{{y}^{2}}}}{2} \\
\end{align}$
Now we have the function as x in terms of y. Now we will try to find the domain of this function. Now the function is only real when $4-{{y}^{2}}\ge 0$
Hence we must have $4>{{y}^{2}}$
Hence the condition on y are $y\in \left[ -2,2 \right]$

Hence the range of the function is $\left[ -2,2 \right]$

Note: Now note that range is all the values of y which a function allows. Hence we can also write the function as y in terms of x and then check what values of y are possible by analyzing the function. Hence we can find the range of the function.