What is the center of the ellipse represented by $\frac{{(x - 6)}^{2}}{36} + \frac{{(y + 4)}^{2}}{16} = 1$ $(x-6)^2/36+(y+4)^2/16=1$ ?

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Answer 1 · 2014-09-28T19:06:54

The center of the ellipse is (6,-4).

$a^{2} = 36$ $a^2=36$
$a = 6$ $a=6$

$b^{2} = 16$ $b^2=16$
$b = 4$ $b=4$

The $x$ $x$ -axis is the major axis . It's vertices/endpoints are $(6 \pm a, - 4)$ $(6+-a,-4)$ .

The $y$ $y$ -axis is the minor axis . It's vertices/endpoints are $(6, - 4 \pm b)$ $(6,-4+-b)$ .

What is the center of the ellipse represented by (x−6)236+(y+4)216=1(x-6)^2/36+(y+4)^2/16=1?