Question

A point moves so that the sum of its distance from the points $(4,0,0)$ and $(-4,0,0)$ remain $10$. The locus of the point is :

• A. $9x^2-25y^2+25z^2=225$
• B. $9x^2-25y^2-25z^2=225$
• C. $9x^2+25y^2+25z^2=225$
• D. $9x^2-25y^2+25z^2+225=0$

Answer 1

The correct option is C $9x^2+25y^2+25z^2=225$

Let the required point be $(x,y,z)$

Given,

Sum of distance of point from $(4,0,0)$ and $(-4,0,0)$ remains 10.

$\sqrt{(x+4{)}^2+y^2+z^2}+\sqrt{(x-4{)}^2+y^2+z^2}=10‘-\left(1\right)$

As we know that

$(x+4{)}^2+y^2+z^2-[(x-4{)}^2+y^2+z^2]=16x-\left(2\right)$

Divide eqs,(2) by (1)-

$\therefore \frac{(x+4{)}^2+y^2+z^2-[(x-4{)}^2+y^2+z^2]}{\sqrt{(x+4{)}^2+y^2+z^2}+\sqrt{(x-4{)}^2+y^2+z^2}}=\frac{16x}{10}$

$⟹\sqrt{(x+4{)}^2+y^2+z^2}-\sqrt{(x-4{)}^2+y^2+z^2}=\frac{8x}{5}-\left(3\right)$

Adding eqs.(1) and (3)

$2\sqrt{(x+4{)}^2+y^2+z^2}=10+\frac{8x}{5}$

Squaring on both sides

$⟹4\left[\right(x+4{)}^2+y^2+z^2]=\frac{2500+64x^2+800x}{25}$

$⟹100(x^2+16+8x+y^2+z^2)=2500+64x^2+800x$

$⟹36x^2+100y^2+100z^2-900=0$

$⟹9x^2+25y^2+25z^2=225$

Hence the locus of the point is $9x^2+25y^2+25z^2=225$