How do you find the coordinates of the center, foci, the length of the major and minor axis given $\frac{{(x + 8)}^{2}}{144} + \frac{{(y - 2)}^{2}}{81} = 1$ $(x+8)^2/144+(y-2)^2/81=1$ ?

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Answer 1 · 2017-05-18T13:18:50

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$\frac{{(x + 8)}^{2}}{12^{2}} + \frac{{(y - 2)}^{2}}{9^{2}} = 1$ $(x+8)^2/(12^2)+(y-2)^2/(9^2)=1$

We compare this equation to the standard equation of a horizontal ellipse

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

The center of the ellipse is $(h, k) = (- 8, 2)$ $(h,k)=(-8,2)$

The length of the major axis is $= 2 a = 2 \cdot 12 = 24$ $=2a=2*12=24$

The length of the major axis is $= 2 b = 2 \cdot 9 = 18$ $=2b=2*9=18$

We need $c$ $c$ to determine the foci

$c^{2} = a^{2} - b^{2} = 144 - 81 = 53$ $c^2=a^2-b^2=144-81=53$

$c = \pm \sqrt{53}$ $c=+-sqrt53$

The foci are $F = (c, 2) = (- 8 + \sqrt{53}, 2)$ $F=(c,2)=(-8+sqrt53,2)$ and

$F'=(-8-sqrt53,2)$
graph{((x+8)^2/144+(y-2)^2/81-1)((x+8)^2+(y-2)^2-0.1)=0 [-26.1, 5.95, -5.07, 10.95]}

How do you find the coordinates of the center, foci, the length of the major and minor axis given (x+8)^2/144+(y-2)^2/81=1(x+8)2144+(y−2)281=1(x+8)^2/144+(y-2)^2/81=1?