Question

Given P= (a,0) and Q= (-a,0) and R is a variable point on one side of the line PQ such that $\angle RPQ-\angle RQP=2\alpha .$ the locus of the point R is

• A. $x^2+y^2+2xycot2\alpha =a^2$
• B. $x^2+y^2+2xytan2\alpha =a^2$
• C. $x^2+y^2-2xytan2\alpha =a^2$
• D. $x^2-y^2+2xycot2\alpha =a^2$

Answer 1

The correct option is D $x^2-y^2+2xycot2\alpha =a^2$

We have,

$In\Delta RMP,$

$\tan \theta =\frac{RM}{MP}=\frac{y_1}{a-x_1}$

$In\Delta RQM,$

$\tan \varphi =\frac{RM}{QM}=\frac{y_1}{a+x_1}$

\end{align}$

And also given that,

$\angle RPQ-\angle RQP=2\alpha$

$\theta -\varphi =2\alpha$

Taking tan both side and we get,

$\tan (\theta -\varphi )=\tan 2\alpha$

$\Rightarrow \frac{\tan \theta -\tan \varphi }{1+\tan \theta \tan \varphi }=\tan 2\alpha$

$\Rightarrow \tan 2\alpha =\frac{\left(\frac{y_1}{a-x_1}\right)-\left(\frac{y_1}{a+x_1}\right)}{1+\left(\frac{y_1}{a-x_1}\right)\left(\frac{y_1}{a+x_1}\right)}$

$\Rightarrow \tan 2\alpha =\frac{\frac{a{y}_1+x_1{y}_1-a{y}_1+x_1{y}_1}{(a-x_1)(a+x_1)}}{\frac{(a-x_1)(a+x_1)+{y_1}^2}{(a-x_1)(a+x_1)}}$

$\Rightarrow \tan 2\alpha =\frac{2x_1{y}_1}{a^2-{x_1}^2+{y_1}^2}$

$\Rightarrow \frac{1}{\cot 2\alpha }=\frac{2x_1{y}_1}{a^2-{x_1}^2+{y_1}^2}$

$\Rightarrow a^2-{x_1}^2+{y_1}^2=2x_1{y}_1\cot 2\alpha$

$\Rightarrow a^2-{x_1}^2+{y_1}^2-2x_1{y}_1\cot 2\alpha =0$

$\Rightarrow {x_1}^2-{y_1}^2+2x_1{y}_1\cot 2\alpha -a^2=0$

Hence, the locus of this equation is

$x^2-y^2+2xy\cot 2\alpha -a^2=0$

$x^2-y^2+2xy\cot 2\alpha =a^2$

Hence, this si the answer.