Locus of the point whose sum of distances from the origin and the x− axis is 4 units is
A
x2=8(2−y),y≥0
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B
x2=8(2+y),y≥0
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C
x2=8(2+y),y<0
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D
x2=8(2−y),y<0
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Solution
The correct option is Cx2=8(2+y),y<0 Let P(x,y) be a point which satisfies the given condition √x2+y2+|y|=4
when y≥0 then √x2+y2=4−y ⇒x2+y2=16−8y+y2 ⇒x2=8(2−y)
when y<0 √x2+y2−y=4 ⇒√x2+y2=4+y ⇒x2+y2=16+8y+y2 ⇒x2=8(2+y)