Question

If $A\left( {5,0} \right)$ and $B\left( { - 5,0} \right)$ are two given points, variable point P such that $PA - PB = 6$. Show that the equation of locus of P is $\dfrac{{{x^2}}}{a} - \dfrac{{{y^2}}}{b} = 1$. Find $a + b$

Answer 1

Hint: This problem deals with finding the locus of an unknown point which is given as a variable point P.
Given the points A and B. Also given that the difference of the distances from point P to the points A to B is 6. Here to find the distance between the points the distance formula is used, when given two points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ , then the distance between these two points is given by:
\[ \Rightarrow \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \]

Complete step-by-step solution:
Given that there are two known points A and B which are given by:
$A = \left( {5,0} \right),B = \left( { - 5,0} \right)$
Let the variable point P be $P\left( {x,y} \right)$
Given that $PA - PB = 6$
Here the distance $PA$ is given by:
$ \Rightarrow PA = \sqrt {{{\left( {x - 5} \right)}^2} + {{\left( {y - 0} \right)}^2}} $
$ \Rightarrow PA = \sqrt {{{\left( {x - 5} \right)}^2} + {y^2}} $
The distance $PB$ is given by:
$ \Rightarrow PB = \sqrt {{{\left( {x + 5} \right)}^2} + {{\left( {y - 0} \right)}^2}} $
$ \Rightarrow PB = \sqrt {{{\left( {x + 5} \right)}^2} + {y^2}} $
Now consider the equation $PA - PB = 6$, which can be re-written as shown below:
$ \Rightarrow PA = 6 + PB$
Substituting the obtained expressions in the above equation, as given below:
$ \Rightarrow \sqrt {{{\left( {x - 5} \right)}^2} + {y^2}} = 6 + \sqrt {{{\left( {x + 5} \right)}^2} + {y^2}} $
Now squaring the above equation on both sides as shown below:
$ \Rightarrow {\left( {\sqrt {{{\left( {x - 5} \right)}^2} + {y^2}} } \right)^2} = 6 + {\left( {\sqrt {{{\left( {x + 5} \right)}^2} + {y^2}} } \right)^2}$
$ \Rightarrow {\left( {x - 5} \right)^2} + {y^2} = {6^2} + {\left( {x + 5} \right)^2} + {y^2} + 2\left( 6 \right)\sqrt {{{\left( {x + 5} \right)}^2} + {y^2}} $
Here expanding the square of the terms and simplifying them further to get simpler expressions:
$ \Rightarrow {x^2} + 25 - 10x + {y^2} = 36 + {x^2} + 25 + 10x + {y^2} + 12\sqrt {{{\left( {x + 5} \right)}^2} + {y^2}} $
$ \Rightarrow 12\sqrt {{{\left( {x + 5} \right)}^2} + {y^2}} = 10x + 36 + 10x$
Grouping the like terms and the unlike terms and simplifying the expressions as given below:
$ \Rightarrow 12\sqrt {{{\left( {x + 5} \right)}^2} + {y^2}} = 20x + 36$
$ \Rightarrow 3\sqrt {{{\left( {x + 5} \right)}^2} + {y^2}} = 5x + 9$
Squaring the above on equation on both sides, as given below:
$ \Rightarrow {\left( {3\sqrt {{{\left( {x + 5} \right)}^2} + {y^2}} } \right)^2} = {\left( {5x + 9} \right)^2}$
$ \Rightarrow 9\left( {{{\left( {x + 5} \right)}^2} + {y^2}} \right) = {\left( {5x + 9} \right)^2}$
Here expanding the square term expression in the above equation, as shown below:
$ \Rightarrow 9{x^2} + 225 + 90x + 9{y^2} = 25{x^2} + 81 + 90x$
$ \Rightarrow 16{x^2} - 9{y^2} = 144$
$ \Rightarrow \dfrac{{16{x^2}}}{{144}} - \dfrac{{9{y^2}}}{{144}} = 1$
$ \Rightarrow \dfrac{{{x^2}}}{9} - \dfrac{{{y^2}}}{{16}} = 1$
Comparing the above equation with $\dfrac{{{x^2}}}{a} - \dfrac{{{y^2}}}{b} = 1$.
Thus $a = 9$ and $b = 16$
$ \Rightarrow a + b = 9 + 16$
$\therefore a + b = 25$

The value of $a + b$ is 25.

Note: Here while solving the problem, please note that when given the equation as $PA - PB = 6$, we should not square the equation directly as the given equation it is, because if we square the equation directly then we would not be able to get the expression of the locus as given, hence migrated the term $PB$ on the other side of the equation.