How do you write an equation of an ellipse given endpoints of major axis at (-11,5) and (7,5) and endpoints of the minor axis at (-2,9) and (-2,1)?

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Answer 1 · 2016-12-12T22:50:04

Please see the explanation for steps leading to the equation.

The endpoints, $(- 11, 5) and (7, 5)$ $(-11, 5) and (7,5)$ , of the major axis (where the x coordinate changes) have a general form of:

$(h - a, k) and (h + a, k)$ $(h - a, k) and (h + a, k)$

This allows us to write the following equations:

$[1]$ $"[1] "$ $k = 5$ $k = 5$
$[2]$ $"[2] "$ $h - a = - 11$ $h - a = -11$
$[3]$ $"[3] "$ $h + a = 7$ $h + a = 7$

The endpoints, $(- 2, 1) and (- 2, 9)$ $(-2, 1) and (-2,9)$ , of the minor axis (where the y coordinate changes) have a general form of:

$(h, k - b) and (h, k + b)$ $(h, k - b) and (h, k + b)$

This allows us to write the following equations:

$[4]$ $"[4] "$ $h = - 2$ $h = -2$
$[5]$ $"[5] "$ $k - b = 1$ $k - b = 1$
$[6]$ $"[6] "$ $k + b = 9$ $k + b = 9$

Subtracting equation [2] from [3] gives us:

$2 a = 18$ $2a = 18$

$a = 9$ $a = 9$

Subtracting equation [5] from [6] gives us:

$2 b = 8$ $2b = 8$

$b = 4$ $b = 4$

All that remains, is to substitute these values into the general form for an ellipse with a horizontal major axis:

$\frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1$ $(x - h)^2/a^2 + (y - k)^2/b^2 = 1$

$\frac{{(x - - 2)}^{2}}{9^{2}} + \frac{{(y - 5)}^{2}}{4^{2}} = 1$ $(x - -2)^2/9^2 + (y - 5)^2/4^2 = 1$