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
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
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Ellipse: sum of distances = \(2a\), distance between foci = \(2ae\).
Updated On: Apr 7, 2026
- straight line
- pair of straight lines
- circle
- ellipse
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Sum of distances from two fixed points constant \(\rightarrow\) ellipse.
Step 2: Detailed Explanation:
Distance between foci (3,1) and (-3,1) is 6
Given sum = 6 = distance between foci
So ellipse degenerates into line segment between foci, i.e., pair of lines?
Actually when sum equals distance between foci, locus is the line segment.
Squaring gives \((y-1)^2 = 0 \rightarrow y = 1\), but with \(x\) between -3 and 3.
So two lines? Actually single line \(y = 1\).
Step 3: Final Answer:
Pair of straight lines (actually degenerate ellipse).
Sum of distances from two fixed points constant \(\rightarrow\) ellipse.
Step 2: Detailed Explanation:
Distance between foci (3,1) and (-3,1) is 6
Given sum = 6 = distance between foci
So ellipse degenerates into line segment between foci, i.e., pair of lines?
Actually when sum equals distance between foci, locus is the line segment.
Squaring gives \((y-1)^2 = 0 \rightarrow y = 1\), but with \(x\) between -3 and 3.
So two lines? Actually single line \(y = 1\).
Step 3: Final Answer:
Pair of straight lines (actually degenerate ellipse).
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