# calculus

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Find the equation of the following. The locus of all points in a plane such that the sum of the distances from points (-2,-1) and (-10,-1) is equal to 6.

• calculus - ,

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

• calculus - ,

Oh shoot. You're right. I took the last part from a different problem. The sum of the distances is equal to 10. Sorry.

• calculus - ,

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
(-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 - ,

Thanks, but where did you get the

2a = 10

from?

• calculus - ,

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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