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:

• A. \( 3x^2 - y^2 = 12 \)
• B. \( x^2 - 3y^2 = 12 \)
• C. \( 4(x^2 - 3y^2) = 1 \)
• D. \( 3x^2 - y^2 = 1 \)

Hint:

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.

Answer 1

The Correct Option is A

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:

\(3x^2 - y^2 = 12\)

Answer 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 \( 3x^2 - y^2 = 12 \).