Question Number 10324 by amir last updated on 04/Feb/17
Answered by mrW1 last updated on 04/Feb/17
$${y}=\frac{\mathrm{1}}{{x}} \\ $$$${slope}\:{of}\:{tangent}\:{line}: \\ $$$${m}_{{t}} \left({x}\right)=\mathrm{tan}\:\theta={y}'\left({x}\right)=−\frac{\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$${let}\:{B}\left({t},{s}\right)\:{be}\:{a}\:{point}\:{on}\:{the}\:{curve}\: \\ $$$${s}=\frac{\mathrm{1}}{{t}} \\ $$$${the}\:{equation}\:{of}\:{the}\:{tangent}\:{line}\:{at} \\ $$$${point}\:{B}\left({t},\frac{\mathrm{1}}{{t}}\right)\:{is} \\ $$$${y}−\frac{\mathrm{1}}{{t}}=−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\left({x}−{t}\right) \\ $$$${its}\:{section}\:{points}\:{on}\:{coordinate}\:{axes}: \\ $$$${C}\left(\mathrm{0},{c}\right)\:{and}\:{D}\left({d},\mathrm{0}\right) \\ $$$$ \\ $$$${c}−\frac{\mathrm{1}}{{t}}=−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\left(\mathrm{0}−{t}\right) \\ $$$$\Rightarrow{c}=\frac{\mathrm{2}}{{t}} \\ $$$$\mathrm{0}−\frac{\mathrm{1}}{{t}}=−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\left({d}−{t}\right) \\ $$$$\Rightarrow{d}=\mathrm{2}{t} \\ $$$$ \\ $$$${the}\:{area}\:{of}\:\Delta{CDO}\:{is}\:{A} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}}×{c}×{d}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}}{{t}}×\mathrm{2}{t}=\mathrm{2}={constant} \\ $$