Find the locus of the point P for which of the distance from P to (4, 0) is double the distance from P to x-axis .

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Solution

Let $P$ be $(x, y)$
Distance from $(4, 0)$
$\sqrt{(x - 4)^{2} + y^{2}} \sqrt{x^{2} - 4 x + 16 + y^{2}} \sqrt{x^{2} + y^{2} - 4 x + 16} \dots (1)$
Distance from Origin
$\sqrt{x^{2} + y^{2}}$
According to Question

$\sqrt{x^{2} + y^{2} - 4 x + 16} = 2 \sqrt{x^{2} + y^{2}}$
Squaring on both sides
$(x^{2} + y^{2} - 4 x + 16) = 4 (x^{2} + y^{2}) 3 x^{2} + 3 y^{2} + 4 x - 16 = 0$

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