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.

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
(-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

Thanks, but where did you get the

2a = 10

from?

by definition, the sum of the two focal length is 2a

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

To find the equation of the locus, we need to understand the definition of a locus. A locus is a set of points that satisfies a given condition or set of conditions. In this case, we are given the condition that the sum of the distances from two points to any point on the locus is equal to 6.

Let's break down the problem into steps to find the equation of the locus:

Step 1: Find the distance between the two given points (-2,-1) and (-10,-1).
We can use the distance formula to find the distance between two points in a plane:
Distance = √((x2 - x1)^2 + (y2 - y1)^2)

Using the formula, we can calculate the distance between (-2,-1) and (-10,-1):
Distance = √((-10 - (-2))^2 + (-1 - (-1))^2) = √((-8)^2 + 0^2) = √(64) = 8

Step 2: Set up the equation representing the sum of the distances from any point on the locus to the two given points (-2,-1) and (-10,-1).
Let's assume the coordinates of a point on the locus are (x,y). According to the problem, the sum of the distances from this point to (-2,-1) and (-10,-1) should be equal to 6. Using the distance formula again, we can set up the equation:
Distance1 + Distance2 = 6

So, the equation becomes:
√((x - (-2))^2 + (y - (-1))^2) + √((x - (-10))^2 + (y - (-1))^2) = 6

Simplifying the equation further:
√((x + 2)^2 + (y + 1)^2) + √((x + 10)^2 + (y + 1)^2) = 6

Therefore, the equation of 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 is √((x + 2)^2 + (y + 1)^2) + √((x + 10)^2 + (y + 1)^2) = 6.