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Let O (0,0) and A (0,1) be two fixed points. Then the locus of a point ‘P’ such that the perimeter of of $Δ A O P$ is 4, is

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$8 x^{2} + 9 y^{2} - 9 y = 18$

$9 x^{2} - 8 y^{2} + 8 y = 16$

$9 x^{2} + 8 y^{2} - 8 y = 16$

$8 x^{2} - 9 y^{2} + 9 y = 18$

answer is A.

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Detailed Solution

$0 = (0, 0), A = (0, 1) & P = (h, k)$

Perimeter of $Δ A O P = 4$

$\Rightarrow A P + O A + O P = 4$

$\Rightarrow A P + O P = 4 - O A$

$\Rightarrow k + 4 = 3 \sqrt{h^{2} + k^{2}}$

S.O.B

$9 h^{2} + 8 K^{2} - 8 K = 16$

Locus of ‘P’ is $9 x^{2} + 8 y^{2} - 8 y = 16$

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