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

Let P be the point ((a/2)(t^2 +(1/t^2 )),a(t−(1/t))) Find the locus of P, as t varies.

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

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

Let P be the point ((a/2)(t^2 +(1/t^2 )),a(t−(1/t))) Find the locus of P, as t varies. Put x=(a/2)((1/t^2 )+t^2 ),y=a(t−(1/t)) ⇒((y/a))^2 =t^2 +(1/t^2 )−2=((2x)/a)−2 ⇒y^2 =2ax−2a^2 ⇒P is on the Parabol y^2 =2ax−2a^2

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

Commented by ZiYangLee last updated on 19/May/21

thank you sir

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

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