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]}

How do you find the foci and sketch the ellipse 9x^2+4y^2=169x^2+4y^2=16?