# Find the equation of the set of points P, the sum of whose distances from A (4, 0, 0)

Question:

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.

Solution:

Let the coordinates of P be (xyz).

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

$\Rightarrow(x-4)^{2}+y^{2}+z^{2}=100-20 \sqrt{(x+4)^{2}+y^{2}+z^{2}+(x+4)^{2}+y^{2}+z^{2}}$

$\Rightarrow x^{2}-8 x+16+y^{2}+z^{2}=100-20 \sqrt{x^{2}+8 x+16+y^{2}+z^{2}}+x^{2}+8 x+16+y^{2}+z^{2}$

$\Rightarrow 20 \sqrt{x^{2}+8 x+16+y^{2}+z^{2}}=100+16 x$

$\Rightarrow 5 \sqrt{x^{2}+8 x+16+y^{2}+z^{2}}=(25+4 x)$

On squaring both sides again, we obtain

$25\left(x^{2}+8 x+16+y^{2}+z^{2}\right)=625+16 x^{2}+200 x$

$\Rightarrow 25 x^{2}+200 x+400+25 y^{2}+25 z^{2}=625+16 x^{2}+200 x$

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

Thus, the required equation is $9 x^{2}+25 y^{2}+25 z^{2}-225=0$.