How do you write an equation of an ellipse in standard form given foci at (- 6, 5) and (- 6, - 7) and whose major axis has length 26?

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Answer 1 · 2017-11-02T15:39:13

The equation is $\frac{{(y + 1)}^{2}}{169} + \frac{{(x + 6)}^{2}}{133} = 1$ $(y+1)^2/169+(x+6)^2/133=1$

The major axis is vertical

The foci are

$F = (- 6, 5) = (x_{0}, y_{0} + c)$ $F=(-6,5)=(x_0,y_0+c)$

$F'=(-6,-7)=(x_0,y_0-c)$

Therefore,

$x_0=-6$

$y_0=-1$

The center of the ellipse is $C=(x_0,y_0)=(-6,-1)$

Also,

$2a=26$ , $=>$ , $a=13$

$FF'=2c=12$ , $=>$ , $c=6$

Therefore,

$b^2=a^2-c^2=13^2-6^2=169-36=133$

$b=sqrt133$

The equation of the ellipse is

$(y-y_0)^2/a^2+(x-x_0)^2/b^2=1$

$(y+1)^2/169+(x+6)^2/133=1$

graph{((y+1)^2/169+(x+6)^2/133-1)(y-1000(x+6))=0 [-36.9, 14.42, -12.38, 13.3]}