Question

Find the relation between x and y when the point Q(x, y) has the property that dist(Q, P1) + dist(Q, P2) = 4 with respect to points P1(0, −1), P2(0, 1).

Answer 1

The relationship between x and y for point Q is determined by the constant sum of distances to points P1(0, −1) and P2(0, 1), which is 4. We use the distance formula to establish an equation and solve for the relationship between x and y.

Step-by-step explanation:

We are given two points, P1(0, −1) and P2(0, 1), and a general point Q(x, y). The relationship between x and y is such that the sum of the distances from Q to P1 and P2 is constant, equal to 4.

To find this relation, we calculate the distances using the distance formula:
d(Q, P1) = √[(x-0)² + (y-(-1))²]
d(Q, P2) = √[(x-0)² + (y-1)²]

Setting the sum of these distances to 4 yields the equation:
√[(x-0)² + (y+1)²] + √[(x-0)² + (y-1)²] = 4.

We have to square this equation twice to eliminate the square roots and obtain a relation between x and y.