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March 26, 2015

March 26, 2015

Posted by **Jamie** on Thursday, April 8, 2010 at 7:35pm.

- calculus -
**Reiny**, Thursday, April 8, 2010 at 9:21pmI think you have an impossible situation

the distance between the two given points is 8 units

The sum of the distance from any point to the two given points has to be greater than 8

- calculus -
**Jamie**, Thursday, April 8, 2010 at 9:46pmOh shoot. You're right. I took the last part from a different problem. The sum of the distances is equal to 10. Sorry.

- calculus -
**Reiny**, Thursday, April 8, 2010 at 10:03pmNow it becomes an ellipse.

from the data

2a = 10

a = 5

the two given points are the foci

and the centre would have to be the midpoint of those, namely

(-6.-1)

the distance form the centre to one focal point is the c value, here it would be 4

In an ellipse with the a horizontal major axis

b^2 + c^2 = a^2

b^2 + 16 = 25

b = 3

in standard form then

(x+6)^2/25 + (y+1)^2/9 = 1

- calculus -
**Jamie**, Thursday, April 8, 2010 at 10:30pmThanks, but where did you get the

2a = 10

from?

- calculus -
**Reiny**, Thursday, April 8, 2010 at 10:42pmby definition, the sum of the two focal length is 2a

check, distance from centre to the x vertex is a + a = 2a

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