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Phoenix [80]
3 years ago
15

The foci of an ellipse are located at (0,6) and (0,-6). The sum of the distances from any point on the ellipse to the foci is 20

. Determine the equation of the ellipse.
Mathematics
1 answer:
zlopas [31]3 years ago
8 0

Answer:

Step-by-step explanation:

We have to sum up some of the major points of an ellipse before we can get to the answer here.  First of all, the foci are ALWAYS on the major axis.  Because we are given the coordinates of the foci as (0, 6) and (0, -6) we know the foci are on the y axis (because x = 0 in our foci coordinates) and the ellipse will have the equation:

\frac{x^2}{b^2} +\frac{y^2}{a^2} =1

This is the equation because: 1. the x and y DO NOT MOVE IN THE EQUATION FOR AN ELLIPSE! THE X WILL ALWAYS BE IN THE FIRST FRACTION! and 2. A is ALWAYS larger than b in an ellipse.  Therefore, if the ellipse is elongated on the y-axis, the a goes under the y axis because the y axis is longer than the x-axis when the foci lie on the the y-axis.  Did I mention that the foci ALWAYS lie on the major axis (the longer one)!? They do!  And the distance from the center to the foci is a length of "c". Ours is 6. Summing up, ours is a vertically stretched ellipse because the foci lie on the y-axis (the vertical axis) and the foci always lie on the major axis.  Hence, the y-axis is the major axis.  And the center is directly in between the 2 foci.  So since our foci are at (0, 6) and (0, -6), our center is at (0, 0).

The vertex also lie on the major axis, at "a".  They are a bit further from the center than the foci are.  The co-vertex is on the minor, or shorter, axis, at "b".  There is a very important thing to know about the foci of an ellipse that will help us solve your problem, which is actually quite a problem indeed!

Pick a point on an ellipse.  The distance from this point to one focal point, let's call that distance d1, when added to the distance from this same point to the other focal point, let's call that distance d2, will ALWAYS BE THE SAME.  Our distances, d1 + d2, we are told equal 20.  

Let's see if we can put some of this together now and solve your problem.  

Because a is the distance from the center to one vertex and a is on the major axis, then the length of the major axis is equal to 2a.

Likewise, because b is the distance from the center to one co-vertex and b is on the minor axis, then the length of the minor axis is equal to 2b.

If d1 + d2 = 20 and d1 + d2 always has the same length as the major axis, then:

2a = 20 and

a = 10.

We already know that c = 6.

To find b, then, we need the equation to find the foci of an ellipse which is

c^2=a^2-b^2

Since we are needing to find the value of b, we will solve this for b-squared:

b^2=a^2-c^2 and since we know a and c:

b^2=10^2-6^2 and

b^2=100-36 and

b^2=64 so

b = 8.  Therefore, the equation of our ellipse, FINALLY, is

\frac{x^2}{64}+\frac{y^2}{100}=1

Yikes!  What a lesson, huh?!

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Plug y=7x+4 into your graphic caculator

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Find the volume of a box with a height that is 4 inches less than the
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Step-by-step explanation:

V = L × W × H

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\bf \stackrel{\textit{cubic feet of air}}{\stackrel{\textit{dependent}}{a(t)}}\qquad \stackrel{\stackrel{\textit{time in minutes}}{independent}}{t}~\hspace{7em}a(t)=6000t


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if we make a table of values for minutes, we know the balloon is inflating at 6000 ft³ each passing minute.

1 minute......................... 6000(1) ft³

2 minutes......................6000(2) ft³

3 minutes......................6000(3) ft³

4 minutes......................6000(4) ft³

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before getting in the domain and range, let's find out how much air after 6 minutes.

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now, what domain, namely values for "t" make sense?

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what values for the range, namely a(t), makes sense?

well, we know when the balloon is empty is holding 0 ft³, and when is full it has 90000 ft³, [0 , 90000].

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