Question 334900
The standard form of the equation of an ellipse is:
{{{(x-h)^2/a^2 + (y-k)^2/b^2 = 1}}}
for horizontally oriented ellipses and
{{{(x-h)^2/b^2 + (y-k)^2/a^2 = 1}}}
for vertically oriented ellipses. In both cases a > b.<br>
Since the origin, (0, 0) is halfway between the two x intercepts and also halfway between the two y intercepts, then the origin is the center of the ellipse. This makes h = 0 and k = 0.<br>
And since a is the distance from the center to the vertices on the major axis and since the distance from (0, 0) to (6, 0) and to (-6, 0) is 6, a = 6 and this ellipse is a horizontally oriented ellipse.<br>
Since b is the distance from the center to the vertices on the minor axis and since the distance from (0, 0) to (0, 4) and to (0, -4) is 4, b = 4.<br>
In summary we have an ellipse that...<ol><li>is horizontally oriented (which means that we will use the first of the two standard forms above).</li><li>has a center at (0, 0) which means that h = 0 and k = 0.</li><li>has a = 6 and b = 4.</li></ol>
Inserting these values for h, k, a and b into the first of the standard forms we get:
{{{(x-0)^2/6^2 + (y-0)^2/4^2 = 1}}}
This simplifies to:
{{{x^2/36 + y^2/16 = 1}}}<br>
The only thing left is to find the foci. The foci are found on the major axis they are a distance of c from the center. The center is (0, 0) and the major axis is horizontal. All we need now is the value of c. We get c by using the equation:
{{{a^2 = b^2 + c^2}}}
Substituting our values for a and b into this equation we can solve for c:
{{{6^2 = 4^2 + c^2}}}
{{{36 = 16 + c^2}}}
{{{20 = c^2}}}
Since we are only interested in the positive value for c we get:
{{{sqrt(20) = c}}}
Simplifying the square root we get:
{{{sqrt(4*5) = c}}}
{{{sqrt(4)*sqrt(5) = c}}}
{{{2sqrt(5) = c}}}
Now we have c. We can use this and the center, (0, 0), to find the coordinates of the foci. Since this is a horizontally oriented ellipse, we will add and subtract c from the x coordinate of the center:
Focus #1: (0+2sqrt(5), 0) or (2sqrt(5), 0)
Focus #2: (0-2sqrt(5), 0) or (-2sqrt(5), 0)