Find the equation of the set of points P,the | vedclass

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Question

(A) Let the coordinates of $P$ be $(x, y, z)$.The coordinates of points $A$ and $B$ are $(4, 0, 0)$ and $(-4, 0, 0)$ respectively.It is given that $PA

Vedclass · Accepted Answer

$9x^{2} + 25y^{2} + 25z^{2} - 225 = 0$

Vedclass · Answer

(A) Let the coordinates of $P$ be $(x, y, z)$.The coordinates of points $A$ and $B$ are $(4, 0, 0)$ and $(-4, 0, 0)$ respectively.It is given that $PA + PB = 10$.$\Rightarrow \sqrt{(x-4)^{2} + y^{2} + z^{2}} + \sqrt{(x+4)^{2} + y^{2} + z^{2}} = 10$.$\Rightarrow \sqrt{(x-4)^{2} + y^{2} + z^{2}} = 10 - \sqrt{(x+4)^{2} + y^{2} + z^{2}}$.On squaring both sides,we obtain:$(x-4)^{2} + y^{2} + z^{2} = 100 - 20\sqrt{(x+4)^{2} + y^{2} + z^{2}} + (x+4)^{2} + y^{2} + z^{2}$.$x^{2} - 8x + 16 + y^{2} + z^{2} = 100 - 20\sqrt{x^{2} + 8x + 16 + y^{2} + z^{2}} + x^{2} + 8x + 16 + y^{2} + z^{2}$.$20\sqrt{x^{2} + 8x + 16 + y^{2} + z^{2}} = 100 + 16x$.$5\sqrt{x^{2} + 8x + 16 + y^{2} + z^{2}} = 25 + 4x$.On squaring both sides again,we obtain:$25(x^{2} + 8x + 16 + y^{2} + z^{2}) = (25 + 4x)^{2}$.$25x^{2} + 200x + 400 + 25y^{2} + 25z^{2} = 625 + 16x^{2} + 200x$.$9x^{2} + 25y^{2} + 25z^{2} - 225 = 0$.Thus,the required equation is $9x^{2} + 25y^{2} + 25z^{2} - 225 = 0$.

Find the equation of the set of points $P$,the sum of whose distances from $A(4, 0, 0)$ and $B(-4, 0, 0)$ is equal to $10$.

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