If the sum of the distances of a point from t | vedclass

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(B) Let the point be $P(x, y)$.The distance of $P$ from the origin $(0, 0)$ is $OP = \sqrt{x^2 + y^2}$.The distance of $P$ from the line $x = 2$ is $|

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$y^2 + 12x = 36$

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(B) Let the point be $P(x, y)$.The distance of $P$ from the origin $(0, 0)$ is $OP = \sqrt{x^2 + y^2}$.The distance of $P$ from the line $x = 2$ is $|x - 2|$.Given that the sum of these distances is $4$,we have $\sqrt{x^2 + y^2} + |x - 2| = 4$.Assuming $x \leq 2$,the equation becomes $\sqrt{x^2 + y^2} = 4 - (2 - x) = 2 + x$ (if $x Considering the standard form,we take $\sqrt{x^2 + y^2} = 6 - x$.Squaring both sides,we get $x^2 + y^2 = (6 - x)^2$.$x^2 + y^2 = 36 + x^2 - 12x$.$y^2 = 36 - 12x$.$y^2 + 12x = 36$.

If the sum of the distances of a point from the origin and the line $x = 2$ is $4$,then its locus is

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