Given the points $A(0,4)$ and $B(0,-4)$ the equation of the locus of the point $P(x,y)$ such that $|AP-BP|=6$ is
$\begin{array}{1 1}(A)\;\frac{y^2}{9}+\frac{x^2}{7}=1\\(B)\;\frac{y^2}{7}+\frac{x^2}{9}=1\\(C)\;\frac{x^2}{9}-1\\(D)\;\frac{y^2}{9} -\frac{x^2}{7}=1 \end{array}$