Question

The locus of a point $P$, such that the sum of whose distances from $A(4,0,0)$ and $B(-4,0,0)$ is equal to $10,$ is

• A. $25x^2+16y^2+9z^2-225=0$
• B. $9x^2+16y^2+25z^2-225=0$
• C. $16x^2+25y^2+25z^2-225=0$
• D. $9x^2+25y^2+25z^2-225=0$

Answer 1

The correct option is D $9x^2+25y^2+25z^2-225=0$

Let the coordinates of $P$ be $(x,y,z).$
The coordinates of points $A$ and $B$ are $(4,0,0)$ and $(-4,0,0)$ respectively.
It is given that $PA+PB=10\cdots \left(i\right)$
Using Distance Formula,$\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$ in equation $\left(i\right)$
$\Rightarrow \sqrt{(x-4{)}^2+(y{)}^2+(z{)}^2}+\sqrt{(x+4{)}^2+(y{)}^2+(z{)}^2}=10$
$\Rightarrow \sqrt{(x-4{)}^2+(y{)}^2+(z{)}^2}=10-\sqrt{(x+4{)}^2+(y{)}^2+(z{)}^2}$
Squaring both sides,
$\Rightarrow (x-4{)}^2+y^2+z^2=100+(x+4{)}^2+(y{)}^2+(z{)}^2-20\sqrt{(x+4{)}^2+y^2+z^2}$
$\Rightarrow x^2+16-8x+y^2+z^2=100+x^2+16+8x+y^2+z^2-20\sqrt{x^2+16+8x+y^2+z^2}$
$\Rightarrow 20\sqrt{x^2+16+8x+y^2+z^2}=100+16x$
$\Rightarrow 5\sqrt{x^2+16+8x+y^2+z^2}=25+4x$
Squaring both sides again,
$\Rightarrow 25(x^2+16+8x+y^2+z^2)=625+16x^2+200x$
$\Rightarrow 25x^2+400+200x+25y^2+25z^2=625+16x^2+200x$
$\Rightarrow 9x^2+25y^2+25z^2-225=0$
Thus the required locus is$9x^2+25y^2+25z^2-225=0$