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Intersection of Tangents to Hyperbola and 1/x

  1. Jun 21, 2012 #1
    1. The problem statement, all variables and given/known data

    ioi4yc.png

    2. Relevant equations



    3. The attempt at a solution

    [itex] \text{Differentiate }{{C}_{2}}:\text{ }\dfrac{d{{y}_{2}}}{dx}=\dfrac{x}{y} [/itex]
    [itex] \text{Therefore, use }Q(-a,-b)\text{ for MQ: }{{y}_{2}}-(-b)=\dfrac{-a}{-b}\left( x-(-a) \right)\text{ }\Leftrightarrow y+b=\dfrac{a}{b}x+\dfrac{{{a}^{2}}}{b} [/itex]
    [itex] \text{Differentiate }{{C}_{1}}:xy=1\Leftrightarrow {{y}_{1}}={{x}^{-1}}=\Rightarrow \dfrac{d{{y}_{1}}}{dx}=-\dfrac{1}{{{x}^{2}}}.
    \text{Therefore, use }P(a,b)\text{ for MP: }{{y}_{1}}-b=-\dfrac{1}{{{a}^{2}}}\left( x-a \right) [/itex]

    [itex]\text{Question asks to prove }x=-b. [/itex]
    [itex] \text{MQ}=MP\Leftrightarrow -b+\dfrac{a}{b}x+\dfrac{{{a}^{2}}}{b}=b-\dfrac{1}{{{a}^{2}}}x+\dfrac{1}{a}\Leftrightarrow x\left( {{a}^{3}}+b \right)=2{{a}^{2}}{{b}^{2}}+ab-{{a}^{4}} [/itex]
    [itex] \Leftrightarrow x\left( {{a}^{3}}+b \right)=2{{\underbrace{\left( ab \right)}_{1}}^{2}}+\underbrace{ab}_{1}-{{a}^{4}}\text{. Now I sub out }a:
    x=\dfrac{2+1-{{a}^{4}}}{{{a}^{3}}+b}=\dfrac{2+1-{{\left( \dfrac{1}{b} \right)}^{4}}}{{{\left( \dfrac{1}{b} \right)}^{3}}+b}=\dfrac{\dfrac{3{{b}^{4}}-1}{{{b}^{4}}}}{\dfrac{1+{{b}^{3}}}{{{b}^{3}}}}=\dfrac{3b^4 - 1}{1 + b^3}.\text{ This is NOT }-b. [/itex]
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
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