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Question Number 141446 by ZiYangLee last updated on 19/May/21

$$\mathrm{Let}\mathrm{P}\mathrm{be}\mathrm{the}\mathrm{point}(\frac{a}{2}(t^2+\frac{1}{t^2}),a(t-\frac{1}{t}))$$$$\mathrm{Find}\mathrm{the}\mathrm{locus}\mathrm{of}\mathrm{P},\mathrm{as}t\mathrm{varies}.$$

Answered by 1549442205PVT last updated on 19/May/21

$$\mathrm{Let}\mathrm{P}\mathrm{be}\mathrm{the}\mathrm{point}(\frac{a}{2}(t^2+\frac{1}{t^2}),a(t-\frac{1}{t}))$$$$\mathrm{Find}\mathrm{the}\mathrm{locus}\mathrm{of}\mathrm{P},\mathrm{as}t\mathrm{varies}.$$$$\mathrm{Put}\mathrm{x}=\frac{\mathrm{a}}{2}(\frac{1}{{\mathrm{t}}^2}+{\mathrm{t}}^2),\mathrm{y}=\mathrm{a}(\mathrm{t}-\frac{1}{\mathrm{t}})$$$$\Rightarrow {\left(\frac{\mathrm{y}}{\mathrm{a}}\right)}^2={\mathrm{t}}^2+\frac{1}{{\mathrm{t}}^2}-2=\frac{2\mathrm{x}}{\mathrm{a}}-2$$$$\Rightarrow {\mathrm{y}}^2=2\mathrm{ax}-{2\mathrm{a}}^2\Rightarrow \mathrm{P}\mathrm{is}\mathrm{on}\mathrm{the}\mathrm{Parabol}$$$${\mathrm{y}}^2=2\mathrm{ax}-{2\mathrm{a}}^2$$

Commented by ZiYangLee last updated on 19/May/21

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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