Question

Given the points A (0, 4) and B (0, -4), the equation of the locus of the point P(x, y) such that |AP-BP|=6 is
(A) $\dfrac{{{x}^{2}}}{7}-\dfrac{{{y}^{2}}}{9}=1$
(B) $\dfrac{{{y}^{2}}}{9}-\dfrac{{{x}^{2}}}{7}=1$
(C) $\dfrac{{{x}^{2}}}{9}-\dfrac{{{y}^{2}}}{7}=1$
(D) $\dfrac{{{y}^{2}}}{7}-\dfrac{{{x}^{2}}}{9}=1$

Answer 1

Hint: For solving this question we will remove modulus from the given equation and then find the locus equation using distance formula. The distance between two points $\left( {{x}_{1}},{{y}_{1}} \right)$and $\left( {{x}_{2}},{{y}_{2}} \right)$is given by the distance formula: $\left( \sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}} \right)$. In an equation or an expression, after modulus function is removed, we will put $\pm $ sign in front of the values, removal of modulus function is done as follows:
$\begin{align}
  & \left| x-y \right|=l \\
 & x-y=\pm l \\
\end{align}$

Complete step by step solution:
As mentioned in the question, we have to find the locus of the point (x, y).
In an equation or an expression, after modulus function is removed, we will put $\pm $ sign in front of the values, removal of modulus function is done as follows:
$\begin{align}
  & \left| x-y \right|=l \\
 & x-y=\pm l \\
\end{align}$
Now, on removing the mod
ulus function from the equation given in the question it will be simplified as follows:
|AP-BP|=6
AP-BP= $\pm $ 6
AP=BP $\pm $ 6
The distance between two points $\left( {{x}_{1}},{{y}_{1}} \right)$and $\left( {{x}_{2}},{{y}_{2}} \right)$is given by the distance formula: $\left( \sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}} \right)$
Now, point A is given as (0, 4) and point B is given as (0, -4).
Here, by using the distance formula we will get:
$\sqrt{{{x}^{2}}+{{\left( y-4 \right)}^{2}}}=\sqrt{{{x}^{2}}+{{\left( y+4 \right)}^{2}}}\pm 6$
(Using the equation that we got after removing the modulus function)
Now, on squaring both side and we get,
$\left[ {{x}^{2}}+{{\left( y-4 \right)}^{2}} \right]={{\left[ \sqrt{{{x}^{2}}+{{\left( y+4 \right)}^{2}}}\pm 6 \right]}^{2}}$
By simplifying this, we get
${{x}^{2}}+{{\left( y-4 \right)}^{2}}=\left( {{x}^{2}}+{{\left( y+4 \right)}^{2}}+36\pm 12\sqrt{\left( {{x}^{2}}+{{\left( y+4 \right)}^{2}} \right)} \right)$
By expanding this,
${{x}^{2}}+{{y}^{2}}+16-8y=\left( {{x}^{2}}+{{y}^{2}}+16+8y+36\pm 12\sqrt{\left( {{x}^{2}}+{{y}^{2}}+16+8y \right)} \right)$
By simplifying this, we will get
$-8y=8y+36\pm 12\sqrt{\left( {{x}^{2}}+{{y}^{2}}+16+8y \right)}$
$\Rightarrow 0=16y+36\pm 12\sqrt{\left( {{x}^{2}}+{{y}^{2}}+16+8y \right)}$
 $\Rightarrow -16x-36=\pm 12\sqrt{\left( {{x}^{2}}+{{y}^{2}}+16+8y \right)}$
By dividing both sides by $4$ we will have
$\Rightarrow -4x-9=\pm 3\sqrt{\left( {{x}^{2}}+{{y}^{2}}+16+8y \right)}$
By squaring on both sides,
$\begin{align}
  & \Rightarrow {{\left( -4y-9 \right)}^{2}}={{\left( \pm 3\sqrt{\left( {{x}^{2}}+{{y}^{2}}+16+8y \right)} \right)}^{2}} \\
 & \Rightarrow 16{{y}^{2}}+81+72y=9\left( {{x}^{2}}+{{y}^{2}}+16+8y \right) \\
\end{align}$
By simplifying this, we will get
$\begin{align}
  & 16{{y}^{2}}+81+72y=9{{x}^{2}}+9{{y}^{2}}+144+72y \\
 & \Rightarrow 7{{y}^{2}}=9{{x}^{2}}+63 \\
 & \Rightarrow 9{{x}^{2}}-7{{y}^{2}}+63=0 \\
\end{align}$
$9{{x}^{2}}-7{{y}^{2}}+63=0$ is the locus of $\left( x,y \right)$.
We can see that the options are not in this form. So, we need to simplify this further.
By further simplifying this we will have
 $\begin{align}
  & 9{{x}^{2}}-7{{y}^{2}}+63=0 \\
 & \Rightarrow 9{{x}^{2}}-7{{y}^{2}}=-63 \\
 & \Rightarrow \dfrac{9{{x}^{2}}-7{{y}^{2}}}{63}=-1 \\
 & \Rightarrow \dfrac{{{x}^{2}}}{7}-\dfrac{{{y}^{2}}}{9}=-1 \\
 & \Rightarrow \dfrac{{{y}^{2}}}{9}-\dfrac{{{x}^{2}}}{7}=1 \\
\end{align}$

So, the correct answer is “Option D”.

Note: While answering questions of this type we can make error while using the distance formulae stated as “The distance between two points $\left( {{x}_{1}},{{y}_{1}} \right)$and $\left( {{x}_{2}},{{y}_{2}} \right)$is given by the distance formula: $\left( \sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}} \right)$” if we forget the distance formulae and take ${{x}_{1}}+{{x}_{2}}$ instead of ${{x}_{1}}-{{x}_{2}}$ and ${{y}_{1}}+{{y}_{2}}$ instead of ${{y}_{1}}-{{y}_{2}}$ which will lead us to a wrong answer. The rule for removing the modulus in an equation or an expression is that we should put $\pm $ sign in front of the values, after the removal of modulus function as follows:
$\begin{align}
  & \left| x-y \right|=l \\
 & x-y=\pm l \\
\end{align}$ .
If we had made a mistake while simplifying this $16{{y}^{2}}+81+72y=9\left( {{x}^{2}}+{{y}^{2}}+16+8y \right)$ and written it as $16{{y}^{2}}+81+72y=9{{x}^{2}}+9{{y}^{2}}+145+72y$ then we will have the simplified equation of locus as $\begin{align}
  & 7{{y}^{2}}=9{{x}^{2}}+64 \\
 & \Rightarrow 9{{x}^{2}}-7{{y}^{2}}+64=0 \\
\end{align}$
which is clearly a wrong answer. So, we should be careful. Some students may try to substitute the coordinates of $A$ and $B$ in the options and check the answer, which is not possible here as the given locus is of $P$ not $A$ and $B$.