Question 233394
<pre><font size = 4 color = "indigo"><b>
It's symmetrical with respect to the x-axis 
because if we replace x by -x and simplify, we
get the same equation we started with:

{{{4x^2+16y^2=16}}}

{{{4(-x)^2+16y^2=16}}}

{{{4x^2+16y^2=16}}}

It's symmetrical with respect to the y-axis also 
because if we replace y by -y and simplify, we
get the same equation we started with:

{{{4x^2+16y^2=16}}}

{{{4x^2+16(-y)^2=16}}}

{{{4x^2+16y^2=16}}}
 
It's symmetrical with respect to the origin 
because if we replace x by -x and y by -y
at the same time and simplify, we also
get the same equation we started with:

{{{4x^2+16y^2=16}}}

{{{4(-x)^2+16(-y)^2=16}}}

{{{4x^2+16y^2=16}}}

Next we find some first quadrant points.

Choose {{{x=0}}}

Substitute in

{{{4x^2+16y^2=16}}}

{{{4(0)^2+16y^2=16}}}
{{{0+16y^2=16}}}
{{{16y^2=16}}}
{{{y^2=1}}}
{{{y^2-1=0}}}
{{{(y-1)(y+1)=0}}}

This gives {{{y=1}}} and {{{y=-1}}}

So a point in the first quadrant is (0,1)

(Actually it's a point on the BORDER of the first
quadrant, on the y-axis)

Choose {{{x=1}}}

Substitute in

{{{4x^2+16y^2=16}}}

{{{4(1)^2+16y^2=16}}}
{{{4+16y^2=16}}}
{{{16y^2=12}}}
{{{y^2=12/16}}}
{{{y^2=3/4}}}

This gives {{{y=sqrt(3)/2}}} and {{{y=-sqrt(3)/2}}}

So a point in the first quadrant is (0,{{{sqrt(3)/2}}}),

which is about (1, .9)

Choose {{{x=2}}}

Substitute in

{{{4x^2+16y^2=16}}}

{{{4(2)^2+16y^2=16}}}
{{{4(4)+16y^2=16}}}
{{{16+16y^2=16}}}
{{{16y^2=0}}}
{{{y^2=0}}}

This gives {{{y=0}}} 

So another point in the first quadrant is (2,0)

(Actually it's another point on the BORDER of the first
quadrant, this time on the x-axis)

Plot those three first quadrant points:

{{{drawing(400,400,-3,3,-3,3,
graph(400,400,-3,3,-3,3),

line(0+.1,1,0-.1,1),line(0,1+.1,0,1-.1),line(0+.1,1+.1,0-.1,1-.1),line(0+.1,1-.1,0-.1,1+.1), 

line(1+.1,sqrt(3)/2,1-.1,sqrt(3)/2),
line(1,sqrt(3)/2+.1,1,sqrt(3)/2-.1),
line(1+.1,sqrt(3)/2+.1,1-.1,sqrt(3)/2-.1),
line(1+.1,sqrt(3)/2-.1,1-.1,sqrt(3)/2+.1),

line(2+.1,0,2-.1,0),line(2,0+.1,2,0-.1),line(2+.1,0+.1,2-.1,0-.1),line(2+.1,0-.1,2-.1,0+.1)



)}}}

Next by the symmetry, we can reflect these three points
in both the x-axis and the y-axis, (like a mirror!) and 
we have this:

{{{drawing(400,400,-3,3,-3,3,
graph(400,400,-3,3,-3,3),

line(0+.1,1,0-.1,1),line(0,1+.1,0,1-.1),line(0+.1,1+.1,0-.1,1-.1),line(0+.1,1-.1,0-.1,1+.1), 

line(1+.1,sqrt(3)/2,1-.1,sqrt(3)/2),
line(1,sqrt(3)/2+.1,1,sqrt(3)/2-.1),
line(1+.1,sqrt(3)/2+.1,1-.1,sqrt(3)/2-.1),
line(1+.1,sqrt(3)/2-.1,1-.1,sqrt(3)/2+.1),

line(2+.1,0,2-.1,0),line(2,0+.1,2,0-.1),line(2+.1,0+.1,2-.1,0-.1),line(2+.1,0-.1,2-.1,0+.1),

line(0+.1,-1,0-.1,-1),line(0,-1+.1,0,-1-.1),line(0+.1,-1+.1,0-.1,-1-.1),line(0+.1,-1-.1,0-.1,-1+.1), 

line(-1+.1,sqrt(3)/2,-1-.1,sqrt(3)/2),
line(-1,sqrt(3)/2+.1,-1,sqrt(3)/2-.1),
line(-1+.1,sqrt(3)/2+.1,-1-.1,sqrt(3)/2-.1),
line(-1+.1,sqrt(3)/2-.1,-1-.1,sqrt(3)/2+.1),

line(-1+.1,-sqrt(3)/2,-1-.1,-sqrt(3)/2),
line(-1,-sqrt(3)/2+.1,-1,-sqrt(3)/2-.1),
line(-1+.1,-sqrt(3)/2+.1,-1-.1,-sqrt(3)/2-.1),
line(-1+.1,-sqrt(3)/2-.1,-1-.1,-sqrt(3)/2+.1),

line(1+.1,-sqrt(3)/2,1-.1,-sqrt(3)/2),
line(1,-sqrt(3)/2+.1,1,-sqrt(3)/2-.1),
line(1+.1,-sqrt(3)/2+.1,1-.1,-sqrt(3)/2-.1),
line(1+.1,-sqrt(3)/2-.1,1-.1,-sqrt(3)/2+.1),


line(-2+.1,0,-2-.1,0),line(-2,0+.1,-2,0-.1),line(-2+.1,0+.1,-2-.1,0-.1),line(-2+.1,0-.1,-2-.1,0+.1)

)}}}

Now we can sketch the curve:

{{{drawing(400,400,-3,3,-3,3,

graph(400,400,-3,3,-3,3, sqrt(16-4x^2)/4 ),
graph(400,400,-3,3,-3,3,-sqrt(16-4x^2)/4 ) )}}}


Edwin</pre></b></font>