If A(2, 3) and B(2, -3) are two points,then t | vedclass

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(A) Let the coordinates of point $P$ be $(x, y)$.Given $PA + PB = 8$.$\sqrt{(x-2)^2 + (y-3)^2} + \sqrt{(x-2)^2 + (y+3)^2} = 8$$\sqrt{(x-2)^2 + (y-3)^2

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$16x^2 + 7y^2 - 64x - 48 = 0$

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(A) Let the coordinates of point $P$ be $(x, y)$.Given $PA + PB = 8$.$\sqrt{(x-2)^2 + (y-3)^2} + \sqrt{(x-2)^2 + (y+3)^2} = 8$$\sqrt{(x-2)^2 + (y-3)^2} = 8 - \sqrt{(x-2)^2 + (y+3)^2}$Squaring both sides:$(x-2)^2 + (y-3)^2 = 64 + (x-2)^2 + (y+3)^2 - 16\sqrt{(x-2)^2 + (y+3)^2}$$y^2 - 6y + 9 = 64 + y^2 + 6y + 9 - 16\sqrt{(x-2)^2 + (y+3)^2}$$-12y - 64 = -16\sqrt{(x-2)^2 + (y+3)^2}$$3y + 16 = 4\sqrt{(x-2)^2 + (y+3)^2}$Squaring again:$9y^2 + 96y + 256 = 16((x-2)^2 + (y+3)^2)$$9y^2 + 96y + 256 = 16(x^2 - 4x + 4 + y^2 + 6y + 9)$$9y^2 + 96y + 256 = 16x^2 - 64x + 16y^2 + 96y + 208$$16x^2 + 7y^2 - 64x - 48 = 0$.

If $A(2, 3)$ and $B(2, -3)$ are two points,then the equation of the locus of a point $P$ such that $PA + PB = 8$ is

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