Question:
If the distance of a variable point \(P\) from a point \(A(2,-2)\) is twice the distance of \(P\) from the Y-axis, then the equation of locus of \(P\) is:
If the distance of a variable point \(P\) from a point \(A(2,-2)\) is twice the distance of \(P\) from the Y-axis, then the equation of locus of \(P\) is:
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Use distance formula and algebraic manipulation to form locus equations.
Updated On: Jul 2, 2025
- \(3x^2 - y^2 + 4x - 4y - 8 = 0\)
- \(x^2 - 4x + 4y + 8 = 0\)
- \(3x^2 - y^2 + 4x - 4y + 8 = 0\)
- \(y^2 - 4x + 4y + 8 = 0\)
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The Correct Option is A
Solution and Explanation
Step 1: Define the distance conditions
Distance from point \(A(2,-2)\) to point \(P(x,y)\) is: \[ d_A = \sqrt{(x-2)^2 + (y+2)^2} \] Distance from point \(P(x,y)\) to Y-axis (x=0) is: \[ d_Y = |x| \] Given, \[ d_A = 2 d_Y \] Step 2: Write the equation
\[ \sqrt{(x-2)^2 + (y+2)^2} = 2 |x| \] Square both sides: \[ (x-2)^2 + (y+2)^2 = 4x^2 \] Expand: \[ x^2 - 4x + 4 + y^2 + 4y + 4 = 4x^2 \] Bring all terms to one side: \[ x^2 - 4x + 4 + y^2 + 4y + 4 - 4x^2 = 0 \] \[ -3x^2 - 4x + y^2 + 4y + 8 = 0 \] Multiply entire equation by \(-1\) to make \(x^2\) positive: \[ 3x^2 + 4x - y^2 - 4y - 8 = 0 \] Rearranged: \[ 3x^2 - y^2 + 4x - 4y - 8 = 0 \]
Distance from point \(A(2,-2)\) to point \(P(x,y)\) is: \[ d_A = \sqrt{(x-2)^2 + (y+2)^2} \] Distance from point \(P(x,y)\) to Y-axis (x=0) is: \[ d_Y = |x| \] Given, \[ d_A = 2 d_Y \] Step 2: Write the equation
\[ \sqrt{(x-2)^2 + (y+2)^2} = 2 |x| \] Square both sides: \[ (x-2)^2 + (y+2)^2 = 4x^2 \] Expand: \[ x^2 - 4x + 4 + y^2 + 4y + 4 = 4x^2 \] Bring all terms to one side: \[ x^2 - 4x + 4 + y^2 + 4y + 4 - 4x^2 = 0 \] \[ -3x^2 - 4x + y^2 + 4y + 8 = 0 \] Multiply entire equation by \(-1\) to make \(x^2\) positive: \[ 3x^2 + 4x - y^2 - 4y - 8 = 0 \] Rearranged: \[ 3x^2 - y^2 + 4x - 4y - 8 = 0 \]
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