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(A) Let the point be $P(x, y)$.According to the problem,the distance of $P$ from $(0, -1)$ is twice its distance from the line $3x + 4y + 1 = 0$.$\sqr

Vedclass · Accepted Answer

$11x^{2} + 39y^{2} + 96xy + 24x - 18y - 21 = 0$

Vedclass · Answer

(A) Let the point be $P(x, y)$.According to the problem,the distance of $P$ from $(0, -1)$ is twice its distance from the line $3x + 4y + 1 = 0$.$\sqrt{(x - 0)^2 + (y + 1)^2} = 2 	imes \frac{|3x + 4y + 1|}{\sqrt{3^2 + 4^2}}$$\sqrt{x^2 + (y + 1)^2} = 2 	imes \frac{|3x + 4y + 1|}{5}$$5\sqrt{x^2 + y^2 + 2y + 1} = 2|3x + 4y + 1|$Squaring both sides:$25(x^2 + y^2 + 2y + 1) = 4(3x + 4y + 1)^2$$25(x^2 + y^2 + 2y + 1) = 4(9x^2 + 16y^2 + 1 + 24xy + 6x + 8y)$$25x^2 + 25y^2 + 50y + 25 = 36x^2 + 64y^2 + 4 + 96xy + 24x + 32y$Rearranging the terms to one side:$(36 - 25)x^2 + (64 - 25)y^2 + 96xy + 24x + (32 - 50)y + (4 - 25) = 0$$11x^2 + 39y^2 + 96xy + 24x - 18y - 21 = 0$

Find the locus of a point such that its distance from the point $(0, -1)$ is twice its distance from the line $3x + 4y + 1 = 0$.

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