Question

Two points A and B have coordinates $(1,0)$ and $( - 1,0)$ respectively and Q is a point which satisfies the relation $AQ - BQ = \pm 1$ . Then find the locus of Q.
A.$12{x^2} - 4{y^2} = 3$
B. $12{x^2} - 4{y^2} + 3 = 3$
C. $12{x^2} + 4{y^2} = 3$
D. $12{x^2} + 4{y^2} + 3 = 3$

Answer 1

Hint: First let the coordinate of Q, then apply the distance formula and obtain the distance between A, Q and B, Q. Substitute the obtained distance in the given equation and calculate to obtain the required locus.

Formula Used:
The distance between the points $(a,b)$ and $(c,d)$ is
$\sqrt {{{\left( {c - a} \right)}^2} + {{\left( {d - b} \right)}^2}} $ .
${(a + b)^2} = {a^2} + 2ab + {b^2}$
${(a - b)^2} = {a^2} - 2ab + {b^2}$
${(a + b + c)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca$

Complete step by step solution:
Suppose the coordinate of Q is $\left( {x,y} \right)$ .
Then,
$AQ = \sqrt {{{(x - 1)}^2} + {{(y - 0)}^2}} $
$ = \sqrt {{{(x - 1)}^2} + {y^2}} $
And,
$BQ = \sqrt {{{(x + 1)}^2} + {{(y - 0)}^2}} $
$ = \sqrt {{{(x + 1)}^2} + {y^2}} $
Substitute $\sqrt {{{(x - 1)}^2} + {y^2}} $for AQ and $\sqrt {{{(x + 1)}^2} + {y^2}} $for BQ in the given equation and calculate to obtain the required result.
$\sqrt {{{(x - 1)}^2} + {y^2}} - \sqrt {{{(x + 1)}^2} + {y^2}} = \pm 1$
Square both sides of the equation,
$\Rightarrow {\left( {\sqrt {{{(x - 1)}^2} + {y^2}} - \sqrt {{{(x + 1)}^2} + {y^2}} } \right)^2} = {\left( { \pm 1} \right)^2}$
$\Rightarrow{\left( {\sqrt {{{(x - 1)}^2} + {y^2}} } \right)^2} + {\left( {\sqrt {{{(x + 1)}^2} + {y^2}} } \right)^2} - 2\sqrt {{{(x - 1)}^2} + {y^2}} \sqrt {{{(x + 1)}^2} + {y^2}} = 1$
$\Rightarrow {(x - 1)^2} + {y^2} + {(x + 1)^2} + {y^2} - 1 = 2\sqrt {{{(x - 1)}^2} + {y^2}} \sqrt {{{(x + 1)}^2} + {y^2}} $
$\Rightarrow {x^2} - 2x + 1 + {y^2} + {x^2} + 2x + 1 + {y^2} - 1 = 2\sqrt {{{(x - 1)}^2} + {y^2}} \sqrt {{{(x + 1)}^2} + {y^2}} $
$\Rightarrow 2{x^2} + 2{y^2} + 1 = 2\sqrt {{{(x - 1)}^2} + {y^2}} \sqrt {{{(x + 1)}^2} + {y^2}} $
Square both sides of the equation,
$\Rightarrow {\left( {2{x^2} + 2{y^2} + 1} \right)^2} = {\left( {2\sqrt {{{(x - 1)}^2} + {y^2}} \sqrt {{{(x + 1)}^2} + {y^2}} } \right)^2}$
$\Rightarrow 4{x^4} + 4{y^4} + 1 + 8{x^2}{y^2} + 4{y^2} + 4{x^2} = 4\left[ {\left( {{{(x - 1)}^2} + {y^2}} \right)\left( {{{(x + 1)}^2} + {y^2}} \right)} \right]$
$\Rightarrow 4{x^4} + 4{y^4} + 1 + 8{x^2}{y^2} + 4{y^2} + 4{x^2} = 4\left[ {\left( {{x^2} + {y^2} - 2x + 1} \right)\left( {{x^2} + {y^2} + 2x + 1} \right)} \right]$
$\Rightarrow 4{x^4} + 4{y^4} + 1 + 8{x^2}{y^2} + 4{y^2} + 4{x^2} = 4\left[ \begin{array}{l}{x^4} + {x^2}{y^2} + 2{x^3} + {x^2} + {x^2}{y^2} + {y^4} + 2x{y^2} + {y^2} - 2{x^3} - 2x{y^2} - 4{x^2}\\ - 2x + {x^2} + {y^2} + 2x + 1\end{array} \right]$
$\Rightarrow 4{x^4} + 4{y^4} + 1 + 8{x^2}{y^2} + 4{y^2} + 4{x^2} = 4\left[ {{x^4} + {y^4} + 2{x^2}{y^2} - 2{x^2} + 2{y^2} + 1} \right]$
$\Rightarrow 4{x^4} + 4{y^4} + 1 + 8{x^2}{y^2} + 4{y^2} + 4{x^2} = 4{x^4} + 4{y^4} + 8{x^2}{y^2} + 8{y^2} - 8{x^2} + 4$
$\Rightarrow 12{x^2} - 4{y^2} = 3$

Option ‘A’ is correct

Note: When solving the given problems like this students often do a common mistake that is they square the equation $2{x^2} + 2{y^2} = 2\sqrt {{{(x - 1)}^2} + {y^2}} \sqrt {{{(x + 1)}^2} + {y^2}} - 1$ without subtracting 1 from each side of the equation and for that reason only they were not able to remove the radical sign. So in this step we need to take the square of the term $2{x^2} + 2{y^2} + 1 = 2\sqrt {{{(x - 1)}^2} + {y^2}} \sqrt {{{(x + 1)}^2} + {y^2}} $ to remove the radical sign otherwise it is not possible to remove the radical sign and get the desired answer.