I 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
Oh shoot. You're right. I took the last part from a different problem. The sum of the distances is equal to 10. Sorry.
Now 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
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
Thanks, but where did you get the
2a = 10
by 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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