Question 124687
<pre><b>
The problem is: 
Sketch a graph of this equation, find the coordinates of the foci, 
and find the lengths of the major and minor axes. 
3x^2+2y^2=24 
I know it may be hard to show how to graph this, so just putting 
it into the correct form of an ellipse, and then helping me figure 
out what the foci and axes are would be very helpful! Thanks :)!!
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We use the fact that the equations of ellipses with centers
at the origin are of two different forms:

{{{x^2/a^2+y^2/b^2=1}}} for ellipses which look like this 

{{{drawing(50,30,-2,2,-1,1,

graph(50,30,-2,2,-1,1, sqrt(4-x^2)/2),

graph(50,30,-2,2,-1,1,-sqrt(4-x^2)/2) )}}},

and have foci (±{{{c}}},0) where {{{c^2=a^2-b^2}}}

and

{{{x^2/b^2+y^2/a^2=1}}} for ellipses which look like this 

{{{drawing(30,50,-1,1,-2,2,

graph(30,50,-1,1,-2,2, sqrt(4-4x^2)),

graph(30,50,-1,1,-2,2,-sqrt(4-4x^2)) )}}}

and have foci (0,±{{{c}}}) where {{{c^2=a^2-b^2}}}

You can tell which is which by looking at
the denominators. It is always true that
{{{a^2}}} is larger than {{{b^2}}}.

{{{a}}} represents the semi-major axis, or as some say,
the "long radius"

{{{b}}} represents the semi-minor axis, or as some say,
the "short radius"

--------

Starting with your equation,

{{{3x^2+2y^2=24}}}

we must get it into one of the forms above.  So
we must first get a 1 on the right by dividing
every term through by 24:

{{{(3x^2)/24+(2y^2)/24=24/24}}}

which simplifies to

{{{x^2/8+y^2/12=1}}}

Since 12 is larger than 8, we know that
{{{12=a^2}}} and {{{8=b^2}}} and so the
ellipse is of the form

{{{ x^2/b^2+y^2/a^2 =1 }}} and looks something
like the second figure above.

Since it has foci (0,±{{{c}}}) where {{{c^2=a^2-b^2}}},
then
{{{c^2=a^2-b^2}}},
{{{c^2=12-8}}}, 
{{{c^2=4}}},
{{{ c=sqrt(4) }}}
{{{ c=2}}}
The foci are 
({{{0}}},±{{{2}}})

Now we'll draw the graph, since {{{aČ=8}}}
Then {{{a=sqrt(12)}}} = {{{sqrt(4*3) = 2sqrt(3)}}},
or about {{{3.5}}} so we draw the major axis
from the vertex point ({{{-3.5}}},{{{0}}}) to the
other vertex point ({{{3.5}}},{{{0}}})

{{{drawing(400,375,-5,5,-5,5,
graph(400,375,-5,5,-5,5),
line(0,-2sqrt(3),0,2sqrt(3)),
rectangle(-.05,-2sqrt(3),.05,2sqrt(3)) )}}}    

Then we draw the minor axis from (-b,0) to (b,0)
and since {{{b^2=8}}}, then {{{b=sqrt(8)}}}, or
{{{b=sqrt(4*2)}}}, or {{{b=2sqrt(2)}}}, or
about {{{2.8}}} 
 so we
draw the minor axis from ({{{-2.8}}},{{{0}}}) to
({{{2.8}}},{{{0}}})

{{{drawing(400,375,-5,5,-5,5,
graph(400,375,-5,5,-5,5),
line(0,-2sqrt(3),0,2sqrt(3)),
rectangle(-.05,-2sqrt(3),.05,2sqrt(3)),
line(-2sqrt(2),0,2sqrt(2),0),
rectangle(-2sqrt(2),-.05,2sqrt(2),.05)
 )}}} 

Now we can sketch in the ellipse and also
indicate the two foci,  
({{{0}}}, ±{{{2}}})
with two short lines

{{{drawing(400,375,-5,5,-5,5,
graph(400,375,-5,5,-5,5,sqrt((24-3x^2)/2)),

line(-.5,2,.5,2),
line(-.5,-2,.5,-2),

graph(400,375,-5,5,-5,5,-sqrt((24-3x^2)/2)),

line(0,-2sqrt(3),0,2sqrt(3)),
rectangle(-.05,-2sqrt(3),.05,2sqrt(3)),

line(-2sqrt(2),0,2sqrt(2),0),
rectangle(-2sqrt(2),-.05,2sqrt(2),.05)
 )}}} 

Edwin</pre>