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  • Two points <img alt="\mathrm{\mathrm{P}(\mathrm{a}, 0) \text { and } \mathrm{Q}(-\mathrm{a}, 0)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cmathrm%7BP%7D%28%5Cmathrm%

Two points \mathrm{\mathrm{P}(\mathrm{a}, 0) \text { and } \mathrm{Q}(-\mathrm{a}, 0)} are given. \mathrm{R} is a variable point on one side of the line \mathrm{PQ} such that \mathrm{\angle RPQ -\angle RQP} is a \mathrm{2\alpha }. Find the locus of \mathrm{R }.

 

Option: 1

\mathrm{x^2+y^2+2 x y\ \cot \alpha-a^2=0}


Option: 2

\mathrm{x^2-y^2-2 x y\ \cot 2 \alpha+a^2=0}


Option: 3

\mathrm{x^2+y^2+2 x y\ \cot 2 \alpha-a^2=0}


Option: 4

\mathrm{x^2-y^2+2 x y\ \cot 2 \alpha-a^2=0}


Answers (1)

best_answer

Let \mathrm{R\left ( h,k \right )} be the variable point (see figure). Then \mathrm{\angle \mathrm{RPQ}=\theta \text { and } \angle \mathrm{RQP}=\varphi \text {, so that } \theta-\varphi=2 \alpha}.
Let \mathrm{RM\perp PQ,} so that \mathrm{RM=k,MP=a-h} and \mathrm{MQ=a+h.}
Then \mathrm{\tan \theta=\frac{\mathrm{RM}}{\mathrm{MP}}=\frac{\mathrm{k}}{\mathrm{a}-\mathrm{h}}=\tan \phi=\frac{\mathrm{RM}}{\mathrm{MQ}}=\frac{\mathrm{k}}{\mathrm{a}+\mathrm{h}}}
Therefore, from \mathrm{2\alpha =\Theta -\varphi ,} we have 
\mathrm{\begin{aligned} & \tan ^2 \alpha=\tan (\theta-\varphi) \\ & =\frac{\tan \theta-\tan \phi}{1+\tan \theta \tan \phi}=\frac{\mathrm{k}(\mathrm{a}+\mathrm{h})-\mathrm{k}(\mathrm{a}-\mathrm{h})}{\mathrm{a}^2-\mathrm{h}^2+\mathrm{k}^2} \\ & \Rightarrow \mathrm{a}^2-\mathrm{h}^2+\mathrm{k}^2 2 \mathrm{hk} \cot 2 \alpha \end{aligned}}
Hence, the locus of \mathrm{\mathrm{R}(\mathrm{h}, \mathrm{k}) \text { is } \mathrm{x}^2-\mathrm{y}^2+2 \mathrm{xy} \cot 2 \alpha-\mathrm{a}^2=0 \text {. }}       
     

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