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Q.
The equation of the set of point $P$, the sum of whose distances from $A(4,0,0)$ and $B(-4,0,0)$ is equal to $10$, is
Introduction to Three Dimensional Geometry
Solution:
Let the point be $P(x, y, z)$. Then, it is given $P A+P B=10$
$ \Rightarrow \sqrt{(x-4)^2+(y-0)^2+(z-0)^2}$
$ +\sqrt{(x+4)^2+(y-0)^2+(z-0)^2}=10 $
$ \Rightarrow \sqrt{(x-4)^2+y^2+z^2}=10-\sqrt{(x+4)^2+y^2+z^2}$
Squaring on both sides, we get
$(x-4)^2+y^2+z^2=100+(x+4)^2+y^2+z^2$
$ -20 \sqrt{(x+4)^2+y^2+z^2} $
$ \Rightarrow x^2+16-8 x=100+x^2+16+8 x$
$ -20 \sqrt{(x+4)^2+y^2+z^2}$
$ \Rightarrow -8 x-8 x-100=-20 \sqrt{(x+4)^2+y^2+z^2}$
$ \Rightarrow -16 x-100=-20 \sqrt{(x+4)^2+y^2+z^2}$
$ \Rightarrow 4 x+25=5 \sqrt{(x+4)^2+y^2+z^2}$
(dividing both sides by $-4$ )
Again, squaring on both sides, we get
$(4 x+25)^2 =25\left[(x+4)^2+y^2+z^2\right]$
$ \Rightarrow 16 x^2+625+200 x =25\left[(x+4)^2+y^2+z^2\right]$
$ \Rightarrow 16 x^2+625+200 x=25\left[x^2+16+8 x+y^2+z^2\right] $
$ \Rightarrow 16 x^2+625+200 x=25 x^2+400+200 x+25 y^2+25 z^2 $
$ \Rightarrow 9 x^2+25 y^2+25 z^2-225=0$
which is the required equation.