AP EAMCET · Maths · Straight Lines
If the distance of a variable point P from a point \(\mathrm{A}(2,-2)\) is twice the distance of P from Y -axis, then the equation of locus of P is
- A \(3 x^2-y^2+4 x-4 y-8=0\)
- B \(x^2-4 x+4 y+8=0\)
- C \(3 x^2-y^2+4 x-4 y+8=0\)
- D \(y^2-4 x+4 y+8=0\)
Answer & Solution
Correct Answer
(A) \(3 x^2-y^2+4 x-4 y-8=0\)
Step-by-step Solution
Detailed explanation
Let the coordinates of point P be \( (x, y) \). Given \( PA = 2 \times (\text{distance of P from Y-axis}) \). \( \sqrt{(x-2)^2 + (y - (-2))^2} = 2|x| \) \( (x-2)^2 + (y+2)^2 = (2x)^2 \) \( x^2 - 4x + 4 + y^2 + 4y + 4 = 4x^2 \) \( 3x^2 - y^2 + 4x - 4y - 8 = 0 \)
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