Draw the graph of the relation `4y^(2)=x^(2)(4-x^(2))`

Explanation

We have `4y^(2)=x^(2)(4-x^(2))`
`therefore y=+-1/2 sqrt(x^(2)(4-x^(2)))`
`therefore y=+-x/2sqrt(4-x^(2))`
Let us first draw the graph of `y=f(x) = x/2sqrt(4-x^(2))`
Clearly, the domain is `[-2.2]`
Now, `f(0)=f(-2)=f(2)=0`
Also for `x lt0, f(x)lt0` and for `x gt0, f(x) gt0`
Now `(dy)/(dx) = 1/2sqrt(4-x^(2))-x/2x/sqrt(4-x^(2))`
`rArr 4-x^(2)=x^(2)`
`rArr x=+-sqrt(2)`
Since `f(x) gt 0` for `x lt 0` and `f(0)=f(2)=0, x=sqrt(2)` is the point of maxima and `x=-sqrt(2)` is the point of minima.
`f(sqrt(2))=1, f(-sqrt(2))=-1`
Also `f(x)` is an odd function.
Hence, the graph of the function is as shown in the following figure.

Graph of `y=-x/1sqrt(4-x^(2))` can be obtained by reflecting the above graph in the x-axis.
Hence, the graph of the relation `4y^(2)=x^(2)(4-x^(2))` is a follows.