Question

Find the locus of a point the sum of whose distance from $(1,0,0)$ and $(-1,0,0)$ is equal to $10$.

• A. $24x^2+25y^2+25z^2-600=0$
• B. $x^2+y^2+z^2-600=0$
• C. $6x^2+5y^2+5z^2-50=0$
• D. $xyz=8$

Answer 1

The correct option is A $24x^2+25y^2+25z^2-600=0$

Let the points $A(1,0,0),B(-1,0,0)$ and $P(x,y,z)$.
Given $PA+PB=0$
$\sqrt{(x-1{)}^2+(y-0{)}^2+(z-0{)}^2}+\sqrt{(x+1{)}^2+(y-0{)}^2+(z-0{)}^2}=10$
$\Rightarrow \sqrt{(x-1{)}^2+y^2+z^2}=10-\sqrt{(x+1{)}^2+y^2+z^2}$
Squaring on both sides, we get
$(x-1{)}^2+y^2+z^2=100+(x+1{)}^2+y^2+z^2-20\sqrt{(x+1{)}^2+y^2+z^2}$
$\Rightarrow -4x-100=-20\sqrt{(x+1{)}^2+y^2+z^2}$
$\Rightarrow x+25=5\sqrt{(x+1{)}^2+y^2+z^2}$
Again squaring on both sides, we get
$x^2+50x+625=25\left\{\right(x^2+2x+1)+y^2+z^2\}$
$\Rightarrow 24x^2+25y^2+25z^2-600=0$
i.e., required equation of locus.