Question

Given that $A=\left( 1,-2 \right)$ & $B=\left( 3,4 \right)$ , find the equation of the Locus of the point P such that, $P{{A}^{2}}+P{{B}^{2}}=A{{B}^{2}}$ .

Answer 1

Hint: For finding the equation of the locus, we have the relation $P{{A}^{2}}+P{{B}^{2}}=A{{B}^{2}}$ . Since, we have already the given coordinates of $A$ and $B$. So, we will consider the coordinates of $P$ . Then, we will find the distance between points $PA$ , $PB$ and $AB$ respectively and will use the value of this distances in the relation $P{{A}^{2}}+P{{B}^{2}}=A{{B}^{2}}$ that will help us to get the equation of the locus.

Complete step by step solution:
Since, we have the relation that is:
$\Rightarrow P{{A}^{2}}+P{{B}^{2}}=A{{B}^{2}}$ … $\left( i \right)$
Since, we have coordinates of point $A$ and $B$ that is $\left( 1,-2 \right)$ and $\left( 3,4 \right)$ respectively. Let us consider that the coordinate of the point $P$ is $\left( x,y \right)$ then the distance between from this point to point $A$ that will be:
$\Rightarrow P{{A}^{2}}={{\left( x-1 \right)}^{2}}+{{\left( y+2 \right)}^{2}}$
Now, we will open the bracket that is in the form of the formula ${{\left( a-b \right)}^{2}}$ and ${{\left( a+b \right)}^{2}}$ . So, we will use this formula as:
$\Rightarrow P{{A}^{2}}={{x}^{2}}+{{1}^{2}}-2\times x\times 1+{{y}^{2}}+{{2}^{2}}+2\times x\times 2$
Now, we will do the calculation required for the above step below as:
$\Rightarrow P{{A}^{2}}={{x}^{2}}+1-2x+{{y}^{2}}+4+4y$
Here, we will combine equal like terms as:
$\Rightarrow P{{A}^{2}}={{x}^{2}}+{{y}^{2}}-2x+4y+4+1$
$\Rightarrow P{{A}^{2}}={{x}^{2}}+{{y}^{2}}-2x+4y+5$
Now, we will find the distance between point $P$ and $B$ as:
$\Rightarrow P{{B}^{2}}={{\left( x-3 \right)}^{2}}+{{\left( y-4 \right)}^{2}}$
Here, we will use the formula ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ as:
$\Rightarrow P{{B}^{2}}={{x}^{2}}+{{3}^{2}}-2\times x\times 3+{{y}^{2}}+{{4}^{2}}-2\times y\times 4$
Now, we will do necessary calculations as:
$\Rightarrow P{{B}^{2}}={{x}^{2}}+9-6x+{{y}^{2}}+16-8y$
Here, we will combine equal like terms as:
$\Rightarrow P{{B}^{2}}={{x}^{2}}+{{y}^{2}}-6x-8y+9+16$
$\Rightarrow P{{B}^{2}}={{x}^{2}}+{{y}^{2}}-6x-8y+25$
Similarly, we will calculate the distance between points $A$ and $B$ as:
$\Rightarrow A{{B}^{2}}={{\left( 3-1 \right)}^{2}}+{{\left( 4+2 \right)}^{2}}$
Here, we will use subtraction and addition as:
$\Rightarrow A{{B}^{2}}={{\left( 2 \right)}^{2}}+{{\left( 6 \right)}^{2}}$
Now, we will do square and will get as:
$\Rightarrow A{{B}^{2}}=4+36$
$\Rightarrow A{{B}^{2}}=40$
Since, we got all the distances. We will use them in the relation mentioned in the equation $\left( i \right)$ as:
$\Rightarrow P{{A}^{2}}+P{{B}^{2}}=A{{B}^{2}}$
$\Rightarrow {{x}^{2}}+{{y}^{2}}-2x+4y+5+{{x}^{2}}+{{y}^{2}}-6x-8y+25=40$
Now, we will simplify the above equation by combining equal like terms as:
$\Rightarrow {{x}^{2}}+{{x}^{2}}+{{y}^{2}}+{{y}^{2}}-2x-6x+4y-8y+5+25=40$
$\Rightarrow 2{{x}^{2}}+2{{y}^{2}}-8x-4y+30=40$
Here, we will subtract $40$ both sides of the above equation as:
$\Rightarrow 2{{x}^{2}}+2{{y}^{2}}-8x-4y+30-40=40-40$
Now, we will have the equation as:
$\Rightarrow 2{{x}^{2}}+2{{y}^{2}}-8x-4y-10=0$
This is the required equation of the locus.

Note: Whenever, we will have to find the equation of the locus. There must be a relation given between points whose equation of locus we need to find. So, with the help of that relation, we can find the equation of locus by getting the distance of the points that are used in the given relation and will put the value in the relation so that we can find the equation of the locus. The formula for distance is:
${{d}^{2}}={{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}$