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Question-10324




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

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