Question

Find the equation of locus of a point, the difference of whose distances from$( -
5,0)$and$(5,0)$is 8.

Answer 1

Hint: Use the distance formula\[d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \]to find
the respective distances between the locus point and the given points$( - 5,0)$and$(5,0)$.
Create an equation by using the difference between these distances which is given as 8.
Simplifying the equation using the identities${(a \pm b)^2} = {a^2} \pm 2ab + {b^2}$so that it is free
from radicals, we get the required equation.

Complete step by step solution:
Given two points$( - 5,0)$and$(5,0)$, and the difference between the distances between the locus and
these points is 8.
Let’s begin by recalling the definition of the equation of a locus.
An equation in 2 variables x and y is said to be the equation of a locus O, if every point of O satisfies the
equation and every point that satisfies the equation of locus belongs to O
Let the point of locus be (x, y).
As the question talks about the distance between points, we will use the distance formula for two points
here.
The distance between two points (x 1 , y 1 ) and (x 2 , y 2 ) is denoted by d and is given by the formula
\[d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \]
Therefore, the distance between (x, y) and (-5, 0) will be
\[
= \sqrt {{{( - 5 - x)}^2} + {{(0 - y)}^2}} \\
= \sqrt {{{( - 5 - x)}^2} + {y^2}} \\
\]
We will use the identities${(a \pm b)^2} = {a^2} \pm 2ab + {b^2}$to simplify further.
Therefore, distance between (x, y) and (-5, 0)
\[ = \sqrt {25 + 10x + {x^2} + {y^2}} \]
Similarly, the distance between (x, y) and (5, 0)
\[
= \sqrt {{{(5 - x)}^2} + {{(0 - y)}^2}} \\
= \sqrt {25 - 10x + {x^2} + {y^2}} \\

\]
According to the given information, we have
\[\sqrt {25 + 10x + {x^2} + {y^2}} - \sqrt {25 - 10x + {x^2} + {y^2}} = 8\]
Let’s simplify this further by taking one of the square roots towards the right hand side of the equation
and then by squaring on both the sides.
Thus, we get
\[\sqrt {25 + 10x + {x^2} + {y^2}} - \sqrt {25 - 10x + {x^2} + {y^2}} = 8\]
\[
\Rightarrow \sqrt {25 + 10x + {x^2} + {y^2}} = \sqrt {25 - 10x + {x^2} + {y^2}} + 8 \\
\Rightarrow 25 + 10x + {x^2} + {y^2} = {(\sqrt {25 - 10x + {x^2} + {y^2}} + 8)^2} \\
\]
Use the identity ${(a \pm b)^2} = {a^2} \pm 2ab + {b^2}$on the RHS again
\[ \Rightarrow 25 + 10x + {x^2} + {y^2} = 25 - 10x + {x^2} + {y^2} + 2\sqrt {(25 - 10x + {x^2} + {y^2})} 8 +
{8^2}\]
Now, eliminate the common terms on both the sides
\[
\Rightarrow 10x = - 10x + 16\sqrt {(25 - 10x + {x^2} + {y^2})} + 64 \\
\Rightarrow 20x - 64 = 16\sqrt {(25 - 10x + {x^2} + {y^2})} \\
\]
Dividing both sides by 4, we get
\[ \Rightarrow 5x - 16 = 4\sqrt {(25 - 10x + {x^2} + {y^2})} \]
Now, square on both the sides
\[
\Rightarrow {(5x - 16)^2} = 16(25 - 10x + {x^2} + {y^2}) \\
\Rightarrow 25{x^2} - 160x + 256 = 400 - 160x + 16{x^2} + 16{y^2} \\
\]
Take the variables on the LHS and the constants on the right
\[
\Rightarrow 25{x^2} - 160x + 256 = 400 - 160x + 16{x^2} + 16{y^2} \\

\Rightarrow 9{x^2} - 16{y^2} = 144 \\
\]
Hence this is the equation of the locus of the point.
Dividing both sides by 144, we get the following equation
\[\dfrac{9}{{144}}{x^2} - \dfrac{{16}}{{144}}{y^2} = 1\]which is the equation of a hyperbola.

Note: Identifying the name of the region defined by the equation gives a deeper geometrical understanding of the concept. From this equation, we see that the hyperbola is centred at origin.