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Question
Find the locus of a point which is equidistant from (1, 3) and x axis.
Solution
Let P(x1, y1) be any point on the locus.
Let A be the point (1, 3)
The distance from the x-axis on the moving pint P(x1, y1) is y1.
Given that AP = y1
AP2 = `y_1^2`
`(x_1 - 1)^2 + (y_1 - 3)^2 = y_1^2`
`x_1^2 - 2x_1 + 1 + y_1^2 - 6y_1 + 9 = y_1^2`
∴ The locus of the point (x1, y1) is x2 – 2x – 6y + 10 = 0
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