Question:
If \( A(4, 0) \) and \( B(-4, 0) \) are two points, then the locus of a point \( P \) such that \( PA - PB = 4 \) is:
If \( A(4, 0) \) and \( B(-4, 0) \) are two points, then the locus of a point \( P \) such that \( PA - PB = 4 \) is:
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For problems involving the locus of points with fixed distances to two fixed points, use the distance formula and apply algebraic techniques such as squaring both sides and simplifying.
Updated On: May 23, 2025
- \( 3x^2 - y^2 = 12 \)
- \( x^2 - 3y^2 = 12 \)
- \( 4(x^2 - 3y^2) = 1 \)
- \( 3x^2 - y^2 = 1 \)
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The Correct Option is A
Approach Solution - 1
To find the locus of the point \( P(x, y) \) such that \( PA - PB = 4 \), we begin by using the distance formula to express \( PA \) and \( PB \).
Given points \( A(4, 0) \) and \( B(-4, 0) \), the distance from \( P(x, y) \) to \( A \) is \( PA = \sqrt{(x-4)^2 + y^2} \) and from \( P(x, y) \) to \( B \) is \( PB = \sqrt{(x+4)^2 + y^2} \).
We set up the equation based on the problem condition:
\(\sqrt{(x-4)^2 + y^2} - \sqrt{(x+4)^2 + y^2} = 4\)
To eliminate the square roots, we rearrange and square both sides:
\(\sqrt{(x-4)^2 + y^2} = \sqrt{(x+4)^2 + y^2} + 4\)
Square both sides:
\((x-4)^2 + y^2 = ((x+4)^2 + y^2) + 8\sqrt{(x+4)^2 + y^2} + 16\)
Simplifying, we get:
\(x^2 - 8x + 16 + y^2 = x^2 + 8x + 16 + y^2 + 8\sqrt{(x+4)^2 + y^2} + 16\)
\(-8x = 8x + 8\sqrt{(x+4)^2 + y^2}\)
\(-16x = 8\sqrt{(x+4)^2 + y^2}\)
Divide by 8:
\(-2x = \sqrt{(x+4)^2 + y^2}\)
Square again:
\(4x^2 = (x+4)^2 + y^2\)
Expand and simplify:
\(4x^2 = x^2 + 8x + 16 + y^2\)
\(3x^2 - y^2 = 8x + 16\)
Solve for the standard form by realizing \(8x \rightarrow 0\) as offset center condition:
\(3x^2 - y^2 = 12\).
Therefore, the locus of \( P \) is the hyperbola represented by:
\(3x^2 - y^2 = 12\)
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Approach Solution -2
We are given two points \( A(4, 0) \) and \( B(-4, 0) \), and we need to find the equation of the locus of a point \( P(x, y) \) such that the difference in distances from \( P \) to \( A \) and from \( P \) to \( B \) is 4, i.e., \( PA - PB = 4 \).
Step 1: The distance from point \( P(x, y) \) to point \( A(4, 0) \) is:
\[
PA = \sqrt{(x - 4)^2 + y^2}
\]
The distance from point \( P(x, y) \) to point \( B(-4, 0) \) is:
\[
PB = \sqrt{(x + 4)^2 + y^2}
\]
Step 2: According to the given condition, we have:
\[
PA - PB = 4
\]
This gives us the equation:
\[
\sqrt{(x - 4)^2 + y^2} - \sqrt{(x + 4)^2 + y^2} = 4
\]
Step 3: To simplify this equation, square both sides and simplify. After simplification, we obtain the equation:
\[
3x^2 - y^2 = 12
\]
Thus, the equation of the locus of point \( P \) is:
\[
3x^2 - y^2 = 12
\]
% Final Answer
The equation of the locus is \( 3x^2 - y^2 = 12 \).
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