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Question

(A) Let the point be $P(h, k)$.According to the given condition,the distance of $P$ from $(4, 0)$ is half of its distance from the line $x - 16 = 0$.S

Vedclass · Accepted Answer

$3x^2 + 4y^2 = 192$

Vedclass · Answer

(A) Let the point be $P(h, k)$.According to the given condition,the distance of $P$ from $(4, 0)$ is half of its distance from the line $x - 16 = 0$.So,$\sqrt{(h - 4)^2 + (k - 0)^2} = \frac{1}{2} |h - 16|$.Squaring both sides,we get $(h - 4)^2 + k^2 = \frac{1}{4} (h - 16)^2$.$4(h^2 - 8h + 16 + k^2) = h^2 - 32h + 256$.$4h^2 - 32h + 64 + 4k^2 = h^2 - 32h + 256$.$3h^2 + 4k^2 = 192$.Replacing $(h, k)$ with $(x, y)$,the locus is $3x^2 + 4y^2 = 192$.

$A$ point moves such that its distance from the point $(4, 0)$ is half that of its distance from the line $x = 16$. The locus of this point is

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