A-variable-point-P-x-y-moves-on-the-curve-xy-6-and-R-is-the-point-which-divides-the-straight-line-that-joins-the-points-A-2-0-and-B-0-1-in-the-ratio-3-1-internally-Find-the-locus-of-PR

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Question Number 124001 by Don08q last updated on 30/Nov/20

$$\mathrm{A}\mathrm{variable}\mathrm{point}P(x,y)\mathrm{moves}\mathrm{on}\mathrm{the}$$$$\mathrm{curve}xy=6\mathrm{and}R\mathrm{is}\mathrm{the}\mathrm{point}\mathrm{which}$$$$\mathrm{divides}\mathrm{the}\mathrm{straight}\mathrm{line}\mathrm{that}\mathrm{joins}\mathrm{the}$$$$\mathrm{points}A(-2,0)\mathrm{and}B(0,-1)\mathrm{in}\mathrm{the}$$$$\mathrm{ratio}3:1,\mathrm{internally}.\mathrm{Find}\mathrm{the}\mathrm{locus}\mathrm{of}$$$$PR.$$

Answered by liberty last updated on 30/Nov/20

$$P(x,\frac{6}{x})\Rightarrow AR:RB=3:1$$$$\Rightarrow \overrightarrow{r}=\frac{\overrightarrow{a}+3\overrightarrow{b}}{1+3}=\left(\begin{array}{c}-\frac{1}{2}\\ 0\end{array}\right)+\left(\begin{array}{c}0\\ -\frac{3}{4}\end{array}\right)=\left(\begin{array}{c}-\frac{1}{2}\\ -\frac{3}{4}\end{array}\right)$$$$R(-\frac{1}{2},-\frac{3}{4}).LetPR=\lambda$$$$\Rightarrow \sqrt{{(x+\frac{1}{2})}^2+{(y+\frac{3}{4})}^2}=\lambda ;$$$$\Rightarrow {(x+\frac{1}{2})}^2+{(y+\frac{3}{4})}^2={\lambda }^2;put\to \{\begin{array}{l}x=1\\ y=6\end{array}$$$$substitute\Rightarrow {(1+\frac{1}{2})}^2+{(6+\frac{3}{4})}^2={\lambda }^2$$$$\Rightarrow \frac{9}{4}+\frac{729}{16}=\frac{36+729}{16}=\frac{765}{16}={\lambda }^2$$$$ThusthelocusofPRis$$$${(x+\frac{1}{2})}^2+{(y+\frac{3}{4})}^2=\frac{765}{16}.$$

Commented by Don08q last updated on 30/Nov/20

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}.$$

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