Let O(0, 0) and A(0, 1) be two fixed points. | vedclass

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(C) Let $P = (h, k)$. The perimeter of $\Delta AOP$ is $AP + OP + AO = 4$.Given $O(0, 0)$ and $A(0, 1)$,we have $AO = 1$.So,$AP + OP = 4 - 1 = 3$.Usin

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

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(C) Let $P = (h, k)$. The perimeter of $\Delta AOP$ is $AP + OP + AO = 4$.Given $O(0, 0)$ and $A(0, 1)$,we have $AO = 1$.So,$AP + OP = 4 - 1 = 3$.Using the distance formula,$\sqrt{h^2 + (k - 1)^2} + \sqrt{h^2 + k^2} = 3$.$\sqrt{h^2 + (k - 1)^2} = 3 - \sqrt{h^2 + k^2}$.Squaring both sides:$h^2 + k^2 - 2k + 1 = 9 + h^2 + k^2 - 6\sqrt{h^2 + k^2}$.$-2k - 8 = -6\sqrt{h^2 + k^2}$.$k + 4 = 3\sqrt{h^2 + k^2}$.Squaring again:$k^2 + 8k + 16 = 9(h^2 + k^2)$.$k^2 + 8k + 16 = 9h^2 + 9k^2$.$9h^2 + 8k^2 - 8k - 16 = 0$.Replacing $(h, k)$ with $(x, y)$,the locus is $9x^2 + 8y^2 - 8y = 16$.

Let $O(0, 0)$ and $A(0, 1)$ be two fixed points. Then the locus of a point $P$ such that the perimeter of $\Delta AOP$ is $4$ is:

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