How do you find the foci and sketch the ellipse #9x^2+4y^2=16#?

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Answer 1 · 2017-05-12T09:40:29

The foci are #F=(0,sqrt20/3)# and #F'=(0,-sqrt20/3)#

The general equation of the ellipse is

#(x-h)^2/a^2+(y-k)^2/b^2=1#

We compare this equation to

#9x^2+4y^2=16#

#9/16x^2+4/16y^2=1#

#x^2/(4/3)^2+y^2/(2^2)=1#

The center of the ellipse is #C=(h,k)=(0,0)# that is, the origin.

The major axis is vertical.

#A=(0,2)# and #A'=(0,-2)#

The other points are

#B=(4/3,0)# and #B'=(-4/3,0)#

We calculate

#c^2=b^2-a^2=4-16/9=20/9#

#c=+-sqrt20/3#

The foci are

#F=(0,sqrt20/3)# and #F'=(0,-sqrt20/3)#

With the points #A#, #A'#, #B# and #B'#, you can sketch the ellipse

graph{(9x^2+4y^2-16)((x)^2+(y-(sqrt20)/3)^2-0.001)((x)^2+(y-sqrt20/3)^2-0.001)=0 [-4.382, 4.386, -2.19, 2.197]}