That's a simple ellipse, the center of which is at (0,0). In addition, the minor and the major axis of this ellipse are on x'x and y'y axons. Because of that, we can do the following:
We can find the points where the ellipse intersects the axons x'x and y'y:
for x'x we put y=0:
9x
2+4*0=36 <=>x
2=4 <=> x=2 or x=-2
So, these points are (2,0) and (-2,0)
for y'y we put x=0:
9*0+4*y
2=36 <=> y
2=9 <=> y=3 or y=-3
So, these points are: (0,3) and (0,-3)
We have the points: (2,0), (-2,0), (3,0), (-3,0)
I always do that, because I rarely remember
the alternative way which is generally prefered:
we have: 9x
2+4y
2=36 <=> x
2/4+y
2/9=1,
or x
2/b
2+y
2/a
2=1
So, b
2=4 <=> b=2 and a
2=9 <=> a=3.
The points where the ellipse intersects the axons are: (b,0), (-b,0), (0,a), (0,-a) which are: (2,0), (-2,0), (3,0), (-3,0)
After that, you can easily sketch the ellipse (see attachment)
The graph was made at:
http://www.quickmath.com/webMathematica3/quickmath/graphs/equations/basic.jsp#c=graph_stepsgraphequation&v1=9x^2%2B4y^2%3D36
As for the foci, you'll have to use the formula:
c
2=|a
2-b
2|. So, c
2=9-4=5 so c=sqrt5.
The foci are on the longer axis. Since y'y axis is longer on our ellipse, the foci should be:
(0,c), (0,-c) or (0,sqrt5), and (0,-sqrt5)
sqrt= square root
(if the y'y axis was longer, we would have: (c,0) and (-c,0) )
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