How do you write an equation of an ellipse given the endpoints of major axis at (10,2) and (-8,2), foci at (6,2) and (-4,2)?

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How do you write an equation of an ellipse given the endpoints of major axis at (10,2) and (-8,2), foci at (6,2) and (-4,2)?

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Answer 1 · 2016-10-28T05:06:01

The equation is $\frac{{(x - 1)}^{2}}{81} + \frac{{(y - 2)}^{2}}{56} = 1$ $((x-1)^2)/81+((y-2)^2)/56=1$

Explanation:

We have here a horitontal major axis:
The center is $(\frac{10 - 8}{2}, 2) = (1, 2)$ $((10-8)/2,2)=(1,2)$
the major axis is $a = 10 - 1 = 9$ $a=10-1=9$
The distance focus to center is $5$ $5$
So we can calculate b from $b^{2} = a^{2} - c^{2}$ $b^2=a^2-c^2$
$b^{2} = 81 - 25 = 56$ $b^2=81-25=56$

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

here $(h, k) = () 1, 2$ $(h,k)=()1,2$ and $a = 9$ $a=9$ And $b = \sqrt{56}$ $b=sqrt56$

So the equation is $\frac{{(x - 1)}^{2}}{81} + \frac{{(y - 2)}^{2}}{56} = 1$ $((x-1)^2)/81+((y-2)^2)/56=1$

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How do you write an equation of an ellipse given the endpoints of major axis at (10,2) and (-8,2), foci at (6,2) and (-4,2)?

1 Answer

Explanation:

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