SOLUTION: find the x-intercept and y-intercept, the foci, the center and circle whether the following equation has a horizonatl or vertical axis: x^2/100 +y^2/36 =1 Please help with thi

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Question 875341: find the x-intercept and y-intercept, the foci, the center and circle whether the following equation has a horizonatl or vertical axis:
x^2/100 +y^2/36 =1
Please help with this!1 Thanks :)
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2%2F100+%2By%5E2%2F36+=1 <----> x%5E2%2F10%5E2+%2By%5E2%2F6%5E2+=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%28horizontal%29 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%28vertical%29 axis of symmetry of the curve.

The x-intercepts, are the points on the x-axis, where y=0 ,
and x%5E2%2F10%5E2+%2B0%5E2%2F6%5E2+=1-->x%5E2%2F10%5E2+=1-->x%5E2=10%5E2-->system%28x=-10%2Cx=610%29 .
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%5E2%2F10%5E2+%2By%5E2%2F6%5E2+=1-->y%5E2%2F6%5E2+=1-->y%5E2=6-->system%28y=-6%2Cy=6%29 .
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%5E2%2Fa%5E2+%2By%5E2%2Fb%5E2+=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 radiusa=b:
x%5E2%2Fa%5E2+%2By%5E2%2Fb%5E2+=1<--->%28x%5E2%2By%5E2%29%2Fb%5E2+=1<--->x%5E2%2By%5E2=b%5E2 ,
but is a%3C%3Eb 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%5E2%2F100+%2By%5E2%2F36+=1 <----> x%5E2%2F10%5E2+%2By%5E2%2F6%5E2+=1 ,
the horizontal span of the ellipse, from (-10,0) to (10,0),
with length 2%2A10=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.

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%5E2=b%5E2%2Bc%5E2 .
So, in this case, 10%5E2=6%5E2%2Bc%5E2
100=36%2Bc%5E2
100-36=c%5E2
64=c%5E2
So, the focal distance is c=sqrt%2864%29 --> highlight%28c=8%29 ,
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).