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. $24x^2+24y^2+25z^2-600=0$
• C. $24x^2+25y^2+24z^2-600=0$
• D. None of these

Answer 1

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

Let the points be $A(1,0,0),B(-1,0,0)$ and $P(x,y,z)$.

Given $PA+PB=10$

$\Rightarrow \sqrt{{(x-1)}^2+{(y-0)}^2+{(z-0)}^2}+\sqrt{{(x+1)}^2+{(y-0)}^2+{(z-0)}^2}=10\Rightarrow \sqrt{{(x-1)}^2+y^2+z^2}=10-\sqrt{{(x+1)}^2+y^2+z^2}$

Squaring both sides, we get

$\Rightarrow {(x-1)}^2+y^2+z^2=100+{(x+1)}^2+y^2+z^2-20\sqrt{{(x+1)}^2+y^2+z^2}\Rightarrow -4x-100=-20\sqrt{{(x+1)}^2+y^2+z^2}\Rightarrow x+25=5\sqrt{{(x+1)}^2+y^2+z^2}$

Again squaring both sides, we get

$x^2+50x+625=25((x^2+2x+1)+y^2+z^2)\Rightarrow 24x^2+25y^2+25z^2-600=0$

which is the required locus.