$\begin{array}{1 1}(A)\;\frac{y^2}{9}+\frac{x^2}{7}=1\\(B)\;\frac{y^2}{7}+\frac{x^2}{9}=1\\(C)\;\frac{x^2}{9}-1\\(D)\;\frac{y^2}{9} -\frac{x^2}{7}=1 \end{array}$

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$|AP-BP|=6$

=>$AP-BP=\pm 6$

=>$ AP=BP\pm 6$

=> $ \sqrt {x^2+(y-4)^2}=\sqrt {x^2+(y+4)^2} \pm 6$

=> $x^2+y^2-8y+16=x^2+y^2+8y+16+36 \pm 12 \sqrt {x^2+(y+4)^2}$

$-16y-36=\pm 12 \sqrt {x^2+(y+4)^2}$

=>$ 4y+9=\pm 3 \sqrt {x^2+(y+4)^2}$

=> $ 16y^2+72y+81=9(x^2+y^2+8y+16)$

=>$7y^2-9x^2=63$

=> $\large\frac{y^2}{9} -\frac{x^2}{7}$$=1$

Hence d is the correct answer.

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