SOLUTION: find the standard form of the equation of the ellipse. vertices: (-4, -1), (-4, 11); endpoints of the minor axis: (-6, 5), (-2, 5)

Question 888991: find the standard form of the equation of the ellipse. vertices: (-4, -1), (-4, 11); endpoints of the minor axis: (-6, 5), (-2, 5)
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
your equation is as follows:

(x+4)^2/4 + (y-6)^2/36 = 1

the graph of your equation is shown below:

look below the graph for further comments.

$$$

the length of your vertical axis is equal to (11 - (-1)) = 12.
the length of your horizontal axis is equal to (-6 - (-2) = 4.

the general form of the equation of an ellipse is:

(x-h)^2 / a^2 + (y-k)^2 / b^2 = 1

a is half the length of the horizontal axis which is equal to 2 which makes a^2 = 4.

b is half the length of the vertical axis which is equal to 6 which makes b^2 = 36

c is the distance between the focal points.

the formula for c is:

c^2 = |a^2 - b^2|

the absolute value sign is necessary because c is always positive and sometimes a is bigger than b (horizontal major axis) and sometimes b is bigger than a (vertical major axis.

your c^2 is equal to |4-25| which is equal to 21 which makes c equal to sqrt(21).

your focal points will be along the major axis at (-4,6-sqrt(21)) and (-4,6+sqrt(21)).

on the graph these focal points show up at (-4,0.417) and (-4,9.583).

the center of your graph is midway between the vertices which puts it at (-4,5).

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