Coordinate Geometry-12

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Find the equation of the locus of the point, such that its distance from the two points (0, a) and (0, -a) are equal.

Solution: Let A be the point (0, a) and B be the point (0, -a). Let P(x, y) be any point on the locus. Therefore, PA = PB

[ √(x - 0)2 + (y - a)2 ] = √[ (x - 0)2 + (y + a)2 ]

Squaring, x2 + (y - a)2 = x2 + (y + a)2

Therefore, (y + a)2 - (y - a)2 = 0 i.e. 4ay = 0

y = 0 which is the x-axis

Therefore, the required locus is the x-axis.

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