Given the points A(0,4) and B(0, -4). Then th | vedclass

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(D) The locus of point $P(x,y)$ is defined by the condition $|AP - BP| = 6$.Here,$A = (0,4)$ and $B = (0,-4)$.$AP = \sqrt{x^2 + (y-4)^2}$ and $BP = \s

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

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(D) The locus of point $P(x,y)$ is defined by the condition $|AP - BP| = 6$.Here,$A = (0,4)$ and $B = (0,-4)$.$AP = \sqrt{x^2 + (y-4)^2}$ and $BP = \sqrt{x^2 + (y+4)^2}$.$|\sqrt{x^2 + (y-4)^2} - \sqrt{x^2 + (y+4)^2}| = 6$.This is the definition of a hyperbola with foci at $(0,4)$ and $(0,-4)$.The distance between foci $2ae = 8$,so $ae = 4$.The constant difference $2a = 6$,so $a = 3$.Since $b^2 = a^2(e^2 - 1) = (ae)^2 - a^2 = 4^2 - 3^2 = 16 - 9 = 7$.The center is at $(0,0)$ and the transverse axis is along the $y$-axis.The equation is $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$,which is $\frac{y^2}{9} - \frac{x^2}{7} = 1$.

Given the points $A(0,4)$ and $B(0, -4)$. Then the equation of the locus of the point $P(x,y)$ such that $|AP - BP| = 6$,is

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