Question 721364
The standard forms for the equation of a ellipse are:
{{{(x-h)^2/a^2+(y-k)^2/b^2=1}}} for horizontal ellipses
and
{{{(x-h)^2/b^2+(y-k)^2/a^2=1}}} for vertical ellipses
In these standard forms...<ul><li>The "h" and "k" are the x and y coordinates of the center of the hyperbola.</li><li>The "a" is the distance from the center to a vertex on the major axis.</li><li>The "b" is the distance from the center to a vertex on the minor axis.</li></ul>
Since we've been told that the major axis is vertical, our ellipse is vertical. So from this point on, we will be using:
{{{(x-h)^2/b^2+(y-k)^2/a^2=1}}}
<ul><li>We know that the center is (-3, 6). So the h and k of the equation will be -3 and 6.</li><li>Since the length of the major axis is 10, the distance from the center to each vertex will be half of that, or 5. So a = 5.</li><li>To find "b" we will use the equation that connects the values of a, b and c in an ellipse:
{{{a^2=b^2+c^2}}}
{{{(5)^2=b^2+(2)^2}}}
{{{25=b^2+4}}}
{{{21=b^2}}}
{{{sqrt(21)=b}}} (We ignore the negative square root since b is a distance.)</li><li>Now that we have the h, k, a and b values we can write the equation:
{{{(x-h)^2/b^2+(y-k)^2/a^2=1}}}<br>
{{{(x-(-3))^2/(sqrt(21))^2+(y-(6))^2/(5)^2=1}}}
which simplifies to:
{{{(x+3)^2/21+(y-6)^2/25=1}}}