<PRE><B>Question 1195932</B>
&lt;br&gt;
{{{3x^2+2y^2=2}}}&lt;br&gt;
The equation has both x^2 and y^2 terms with both positive, so the conic is an ellipse.  There are no linear terms, so the center of the ellipse is at the origin.  The standard form is either {{{x^2/a^2+y^2/b^2=1}}} if the major axis is horizontal or {{{x^2/b^2+y^2/a^2=1}}}  if the major axis is vertical.  (For an ellipse, a &gt; b, so a^2 &gt; b^2.)&lt;br&gt;
To put the given equation in standard form, divide the whole equation by 2:&lt;br&gt;
{{{(3/2)x^2+y^2=1}}}&lt;br&gt;
then put the left-hand side in the required form:&lt;br&gt;
{{{x^2/(2/3)+y^2/1=1}}}&lt;br&gt;
Because 1 &gt; 2/3, the major axis is vertical, so the x coordinates of the foci are still 0.  That eliminates answer choices A and C.&lt;br&gt;
The distance from the center to each focus is c, where a, b, and c are related by&lt;br&gt;
{{{c^2=a^2-b^2}}}&lt;br&gt;
So&lt;br&gt;
{{{c^2=1-2/3=1/3}}}
{{{c=sqrt(1/3)}}}&lt;br&gt;
So the two foci are a distance {{{sqrt(1/3)=1/sqrt(3)}}} above and below the center (0,0), giving&lt;br&gt;
ANSWER: D&lt;br&gt;
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