Question

Find the equation of the set point $P$, the sum of whose distance from $A\left( {4,0,0} \right)$ and $B\left( { - 4,0,0} \right)$ is equal to 10.

Answer 1

Let the point $P$ be $\left( {x,y,z} \right)$. Then, we will calculate distance from $P\left( {x,y,z} \right)$ to $A\left( {4,0,0} \right)$ and from $P\left( {x,y,z} \right)$ to $B\left( { - 4,0,0} \right)$ using distance formula. Next, substitute the values in the expression $PA + PB = 10$ and then simplify the equation to find an equation of the set of points of $P$.

Complete step-by-step answer:
Let the coordinates of point $P$ be $\left( {x,y,z} \right)$
Then, according to the given conditions, we have $PA + PB = 10$
Now, we know that distance between 2 points $\left( {{x_1},{y_1},{z_1}} \right)$ and $\left( {{x_2},{y_2},{z_2}} \right)$ is given as $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}}$
Then, the distance from $P\left( {x,y,z} \right)$ to $A\left( {4,0,0} \right)$ will be
$\Rightarrow$ $\sqrt {{{\left( {x - 4} \right)}^2} + {{\left( {y - 0} \right)}^2} + {{\left( {z - 0} \right)}^2}} = \sqrt {{{\left( {x - 4} \right)}^2} + {{\left( y \right)}^2} + {{\left( z \right)}^2}}$
The distance from $P\left( {x,y,z} \right)$ to $B\left( { - 4,0,0} \right)$
$\Rightarrow$ $\sqrt {{{\left( {x - \left( { - 4} \right)} \right)}^2} + {{\left( {y - 0} \right)}^2} + {{\left( {z - 0} \right)}^2}} = \sqrt {{{\left( {x + 4} \right)}^2} + {{\left( y \right)}^2} + {{\left( z \right)}^2}}$
We will substitute these values in the relation $PA + PB = 10$
$\Rightarrow$ $\sqrt {{{\left( {x - 4} \right)}^2} + {{\left( y \right)}^2} + {{\left( z \right)}^2}} + \sqrt {{{\left( {x + 4} \right)}^2} + {{\left( y \right)}^2} + {{\left( z \right)}^2}} = 10$
We will take any one of the terms from LHS to RHS.
$\Rightarrow$ $\sqrt {{{\left( {x - 4} \right)}^2} + {{\left( y \right)}^2} + {{\left( z \right)}^2}} = 10 - \left( {\sqrt {{{\left( {x + 4} \right)}^2} + {{\left( y \right)}^2} + {{\left( z \right)}^2}} } \right)$
Now, take square on both sides and apply the formula, {\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$$$ {\left( {x - 4} \right)^2} + {\left( y \right)^2} + {\left( z \right)^2} = 100 + {\left( {x + 4} \right)^2} + {\left( y \right)^2} + {\left( z \right)^2} - \left( {20\sqrt {{{\left( {x + 4} \right)}^2} + {{\left( y \right)}^2} + {{\left( z \right)}^2}} } \right) \\\ \Rightarrow {\left( {x - 4} \right)^2} = 100 + {\left( {x + 4} \right)^2} - \left( {20\sqrt {{{\left( {x + 4} \right)}^2} + {{\left( y \right)}^2} + {{\left( z \right)}^2}} } \right) \\\ $ Apply the formulas, {\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$and${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab $ {x^2} + 16 - 8x = 100 + {x^2} + 16 + 8x - \left( {20\sqrt {{{\left( {x + 4} \right)}^2} + {{\left( y \right)}^2} + {{\left( z \right)}^2}} } \right) \\\ 100 - 16x = \left( {20\sqrt {{{\left( {x + 4} \right)}^2} + {{\left( y \right)}^2} + {{\left( z \right)}^2}} } \right) \\\ $ We can take 4 common from both LHS and RHS 25 - 4x = \left( {5\sqrt {{{\left( {x + 4} \right)}^2} + {{\left( y \right)}^2} + {{\left( z \right)}^2}} } \right) Again take square on both sides, $ 625 + 16{x^2} + 200x = 25\left( {{{\left( {x + 4} \right)}^2} + {{\left( y \right)}^2} + {{\left( z \right)}^2}} \right) \\\ \Rightarrow 625 + 16{x^2} + 200x = 25\left( {{x^2} + 16 + 8x + {y^2} + {z^2}} \right) \\\ \Rightarrow 9{x^2} + 25{y^2} + 25{z^2} = 225 \\\ $ Thus, the equation of the set of point $P$ is9{x^2} + 25{y^2} + 25{z^2} = 225$$

Note: The distance formula between two points $\left( {{x_1},{y_1},{z_1}} \right)$ and $\left( {{x_2},{y_2},{z_2}} \right)$ is given as $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}}$. When we have two terms containing variables and square root on one side, then we bring one to the other side and then take square root to avoid unnecessary calculations.