Question 875341
{{{x^2/100 +y^2/36 =1}}} <----> {{{x^2/10^2 +y^2/6^2 =1}}}
 
Since {{{y}}} appears only as its square,
substituting  {{{-y}}} for {{{y}}} gives us the same equation, meaning that the curve is symmetrical to both sides (above and below) the x-axis.
So, the x-axis is a {{{highlight(horizontal)}}} axis of symmetry of the curve
Since {{{x}}} appears only as its square,
substituting  {{{-x}}} for {{{x}}} gives us the same equation, meaning that the curve is symmetrical to both sides (left and right)of the y-axis.
So, the y-axis is a {{{highlight(vertical)}}} axis of symmetry of the curve.
 
The x-intercepts, are the points on the x-axis, where {{{y=0}}} ,
and {{{x^2/10^2 +0^2/6^2 =1}}}-->{{{x^2/10^2 =1}}}-->{{{x^2=10^2}}}-->{{{system(x=-10,x=610)}}} .
So the x-intercepts are the points (-10,0) and (10,0).
The y-intercepts, are the points on the y-axis, where {{{x=0}}} ,
and {{{0^2/10^2 +y^2/6^2 =1}}}-->{{{y^2/6^2 =1}}}-->{{{y^2=6}}}-->{{{system(y=-6,y=6)}}} .
So the y-intercepts are the points (0,-6) and (0,6).
With the x- and y-axes as axes of symmetry, we get the idea that the curve has the origin as its center.
 
GENERAL INFORMATION:
The equation {{{x^2/a^2 +y^2/b^2 =1}}} is the equation of an ellipse,
centered at the origin, with the x- and y-axes for axes of symmetry.
If we make {{{a=b}}} it turns into s circle with radius{{{a=b}}}:
{{{x^2/a^2 +y^2/b^2 =1}}}<--->{{{(x^2+y^2)/b^2 =1}}}<--->{{{x^2+y^2=b^2}}} ,
but is {{{a<>b}}} it will be stretched (horizontally or vertically) into an ellipse.
The ellipse span along the x-axis measures {{{2a}}} (with extreme points at (-a,0) and (a,0)),
and along the y-axis the ellipse measures {{{2b}}} (with extreme points at (0,-b) and (0,b)).
 
In the case of {{{x^2/100 +y^2/36 =1}}} <----> {{{x^2/10^2 +y^2/6^2 =1}}} ,
the horizontal span of the ellipse, from (-10,0) to (10,0),
with length {{{2*10=20}}} , is longer than the vertical span, from(0,-6) to (0,6),
so the portion of the horizontal axis connecting vertices (-10,0) and (10,0) is called the major axis.
The segment of the vertical axis ending in (0,-6) and (0,6) is called the minor axis, and its ends may be called co-vertices of the ellipse.
{{{drawing(400,300,-12,12,-9,9,
grid(1),arc(0,0,20,12,0,360),
green(line(-10,0,10,0)),
blue(line(0,-6,0,6))
)}}}
The distance from center to the vertices {{{a=10}}} is called the semi-major axis,
and the distance from center to the co-vertices {{{b=6}}} is called the semi-minor axis.
The foci are at a distance {{{c}}} from the center of the ellipse, and
{{{a^2=b^2+c^2}}} .
So, in this case, {{{10^2=6^2+c^2}}}
{{{100=36+c^2}}}
{{{100-36=c^2}}}
{{{64=c^2}}}
So, the focal distance is {{{c=sqrt(64)}}} --> {{{highlight(c=8)}}} ,
and the foci are at that distance from the center, along the major axis.
That means that the foci are the points (-8,0) and (8,0).
{{{drawing(400,300,-12,12,-9,9,
grid(1),arc(0,0,20,12,0,360),
green(line(-10,0,10,0)),
blue(line(0,-6,0,6)),
red(line(8,0,0,6)),red(line(-8,0,0,6)),
circle(-8,0,0.3),circle(8,0,0.3)
)}}}