Question

Let $P=\left(1,0\right)$ and $Q=\left(- 1,0\right)$ are two fixed points and $R$ is a variable point on one side of the line $PQ$ such that $\angle RPQ-\angle RQP=45^{o},$ then the locus of $R$ is

• A. $y^{2}-x^{2}+2xy-1=0$
• B. $x^{2}-y^{2}+2xy+1=0$
• C. $y^{2}+x^{2}-2xy=1$
• D. $y^{2}-x^{2}-2xy+1=0$

Answer 1

Answer: (D)

Let, $\angle RPQ=\theta $ & $\angle RQP=\phi$ , where $\theta -\phi=45^{^\circ }$
Now, $tan\left(\theta - \phi\right)=1$
$\Rightarrow \frac{tan \theta - tan ⁡ \phi}{1 + tan ⁡ \theta tan ⁡ \phi}=1$ , where $tan \theta =\frac{y}{1 - x}\&tan⁡\phi=\frac{y}{x + 1}$
$\Rightarrow \frac{y}{1 - x}-\frac{y}{x + 1}=1+\left(\frac{y}{1 - x}\right)\left(\frac{y}{x + 1}\right)$
$\Rightarrow 2yx=1-x^{2}+y^{2}$
$\Rightarrow y^{2}-x^{2}-2xy+1=0$