Question 1195932: This is a question involving conic sections. Can someone explain #14 on https://www.math.purdue.edu/php-scripts/courses/oldexams/serve_file.php?file=16200FE-F2018.pdf ?
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
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  if the major axis is horizontal or  if the major axis is vertical. (For an ellipse, a > b, so a^2 > b^2.)
To put the given equation in standard form, divide the whole equation by 2:
then put the left-hand side in the required form:
Because 1 > 2/3, the major axis is vertical, so the x coordinates of the foci are still 0. That eliminates answer choices A and C.
The distance from the center to each focus is c, where a, b, and c are related by
So
So the two foci are a distance  above and below the center (0,0), giving
ANSWER: D
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