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

x^{2} + 3 y^{2} + 8 x + 16 = 0

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x^{2} + 3 y^{2} - 8 x + 16 = 0

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x^{2} - 3 y^{2} + 8 x - 16 = 0

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x^{2} - 3 y^{2} + 8 x + 16 = 0

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Solution

The correct option is D $x^{2} - 3 y^{2} + 8 x + 16 = 0$
$(h + 4)^{2} + (k - 0)^{2} = (2 k)^{2}$
$h^{2} - 3 k^{2} + 8 h + 16 = 0$
$x^{2} - 3 y^{2} + 8 x + 16 = 0$

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Similar questions

Write $x^{2}$ +8x+16 =0 in the form $x^{2}$ +px+qx+16 =0 such that p x q = 16

x^{2} + 8 x + 16 = 0

x^{2} + 8 x + 16 = 0

x^{2} - 8 x + 16 = 0

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