Question

Find the locus of the point, the sum of whose distances from the points A(4, 0, 0) and B(-4, 0, 0) is equal to 10.

Answer 1

Let locus of P$(x_1,y_1,z)$ is the required locus, so

AP + BP = 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^{2}+16-8x+y^{2}+z^{2}}=10-

$\sqrt{x^2+8x+16+y^2+z^2}$

$\Rightarrow x^2+y^2+z^2-8x+16=(10{)}^2+(x^2+y^2+z^2+8x+16)-20\sqrt{x^2+y^2+z^2+8x+16}$

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

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

$\Rightarrow -4(4x+25)=-20\sqrt{x^2+y^2+z^2+8x+16}$

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

Squaring both the sides,

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

$\Rightarrow 16x^2+625+200x=25(x^2+y^2+z^2+8x+16)$

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

$\Rightarrow 9x^2+25y^2+25z^2-225=0$ is the required locus.