Question 1041900
<pre><b>
We get it into one of the forms:


{{{(x-h)^2/a^2+(y-k)^2/b^2 = 1}}} or {{{(x-h)^2/b^2+(y-k)^2/a^2 = 1}}}

2x²+8x+3y²-12y = 10

2(x²+4x)+3(y²-4y) = 10

Take half of 4, get 2.
Square 2, get 4.
Add 4 inside the first parentheses. 
Since the first parentheses is 
multiplied by 2, adding 2 inside
the first parentheses amounts to
adding 2 times 4, or 8 to the left
side.  So add 8 to the right side
to offset.

2(x²+4x+4)+3(y²-4y) = 10+8

Take half of -4, get -2.
Square -2, get 4.
Add 4 inside the second parentheses. 
Since the second parentheses is 
multiplied by 3, adding 4 inside
the second parentheses amounts to
adding 3 times 4, or 12 to the left
side.  So add 12 to the right side
to offset.

2(x²+4x+4)+3(y²-4y+4) = 10+8+12

Factor each parenthetical expression
as the square of a binomial and collect
terms on the right.

2(x+2)²+3(y-2)² = 30

Get 1 on the right side by dividing
every term by 30

{{{2(x+2)^2/30+3(y-2)^2/30 = 30/30}}}

{{{(x+2)^2/15+(y-2)^2/10 = 1}}}

Since the larger denominator is under the 
term in x, we can tell that the ellipse 
looks like this {{{drawing(20,10,-2,2,-1,1,arc(0,0,-3.9,1.9) )}}}
 and not like this {{{drawing(10,20,-1,1,-2,2,arc(0,0,1.9,-3.9) )}}}.

We compare it to the standard form for an ellipse:

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

h=-2, k=2, a²=15, b²=10
So the center is (h,k) = (-2,2) and

{{{a=sqrt(15)}}}, {{{b=sqrt(10)}}}

Those are the semi-major and semi-minor axes. You
call them long and short radii?

{{{drawing(400,320,-7,3,-2,6, graph(400,320,-7,3,-2,6),
arc(-2,2,2sqrt(15),2sqrt(10)),
green(line(-2,2,-2,2+sqrt(10)),line(-2,2,-2+sqrt(15),2)),
locate(-3.3,2.4,"(-2,2)"), green(locate(-1.85,4,sqrt(10)),
locate(-1.2,1.9,sqrt(15)))   


)}}}

Edwin</pre></b>