Question

The locus of the point \(P(x, y)\) satisfying \(\sqrt{(x-3)^2 + (y-1)^2} + \sqrt{(x+3)^2 + (y-1)^2} = 6\) is

• A. straight line
• B. pair of straight lines
• C. circle
• D. ellipse

Hint:

Ellipse: sum of distances = \(2a\), distance between foci = \(2ae\).

Answer 1

The Correct Option is B

To determine the locus of the point \( P(x, y) \) satisfying the condition:

\(\sqrt{(x-3)^2 + (y-1)^2} + \sqrt{(x+3)^2 + (y-1)^2} = 6\)

This equation is of the form:

\(\sqrt{(x-x_1)^2 + (y-y_1)^2} + \sqrt{(x-x_2)^2 + (y-y_2)^2} = 2a\)

where \( (x_1, y_1) \) and \( (x_2, y_2) \) are the foci of an ellipse, and \( 2a \) is the constant sum of distances.

Here, \( (x_1, y_1) = (3, 1) \), \( (x_2, y_2) = (-3, 1) \) and \( 2a = 6 \), which gives \( a = 3 \).

The distance between the foci:

\(\sqrt{(3 - (-3))^2 + (1 - 1)^2} = \sqrt{6^2} = 6\)

Since \( 2a = 6 \) equals the distance between the two foci, this condition represents a degenerate ellipse, which results in a pair of straight lines.

Therefore, the locus of the point \( P(x, y) \) is a pair of straight lines.

Hence, the correct answer is: pair of straight lines.