Question 612377
I assume the problem is to find the equation of the conic section being described. In the future, please include the directions for problem.<br>
With a major axis and two foci we are dealing with either an ellipse or a hyperbola.<br>
Since the major axis is 16 units long, "a" must be 8.<br>
Since the foci are (0, {{{2sqrt(15)}}}) and (0, {{{-2sqrt(15)}}}) and since the center of an ellipse or hyperbola is always half way between the foci, the center must be at (0, 0)<br>
With foci of (0, {{{2sqrt(15)}}}) and (0, {{{-2sqrt(15)}}}) and with a center of (0, 0), the "c" must be {{{2sqrt(15)}}}<br>
Since {{{sqrt(15) < 4}}}, {{{2*sqrt(15) < 8}}}. So c < 8. And since c < a we must be dealing with an ellipse.<br>
Since the foci are up and down from each other (the x's are the same but the y's are different), the major axis of the ellipse is vertical.<br>
The only piece of information we still need is the "b". The relationship between a, b and c in an ellipse is:
{{{a^2 = b^2 + c^2}}}
Substituting in our values for a and c we get:
{{{(8)^2 = b^2 + (2sqrt(15))^2}}}
Simplifying:
{{{64 = b^2 + 4*15}}}
{{{64 = b^2 + 60}}}
Subtracting 60:
{{{4 = b^2}}}
2 = b (We can ignore the fact that b could also be -2 since b must be positive.)<br>
The standard form for the equation of a vertically-oriented ellipse is:
{{{(x-h)^2/b^2 + (y-k)^2/a^2 = 1}}}
with "h" being the x coordinate of the center and "k" being the y coordinate of the center. Substituting in our center and "a" and "b" we get:
{{{(x-0)^2/2^2 + (y-0)^2/8^2 = 1}}}
which simplifies to:
{{{x^2/4 + y^2/64 = 1}}}