The locus of a point P, whose distance from the point (1, -2) is always two times its distance from y-axis is.

x^{2} + y^{2} + 2 x - 4 y - 5 = 0

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x^{2} - y^{2} - 2 x + 4 y + 5 = 0

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3 x^{2} - y^{2} + 2 x - 4 y - 5 = 0

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3 x^{2} + y^{2} + 2 x - 4 y - 5 = 0

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Solution

The correct option is C $3 x^{2} - y^{2} + 2 x - 4 y - 5 = 0$
Let any point on the curve be $(x, y)$ . Then distance between the point and y axis
$d_{y} = | x |$ ...(i)
Distance from the point $(1, - 2)$ is
$d = \sqrt{(x - 1)^{2} + (y + 2)^{2}}$ ...(ii)
Now,
$d = 2 d_{y}$
$d^{2} = 4 d_{y}^{2}$
$(x - 1)^{2} + (y + 2)^{2} = 4 x^{2}$
$x^{2} - 2 x + 1 + y^{2} + 4 y + 4 = 4 x^{2}$
$3 x^{2} - y^{2} + 2 x - 4 y - 5 = 0$
It is the required equation of the curve.

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