Intersection of Tangents to Hyperbola and 1/x

[12john]

Homework Statement

Homework Equations

The Attempt at a Solution

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

$\text{Question asks to prove }x=-b.$
$\text{MQ}=MP\Leftrightarrow -b+\dfrac{a}{b}x+\dfrac{{{a}^{2}}}{b}=b-\dfrac{1}{{{a}^{2}}}x+\dfrac{1}{a}\Leftrightarrow x\left( {{a}^{3}}+b \right)=2{{a}^{2}}{{b}^{2}}+ab-{{a}^{4}}$
$\Leftrightarrow x\left( {{a}^{3}}+b \right)=2{{\underbrace{\left( ab \right)}_{1}}^{2}}+\underbrace{ab}_{1}-{{a}^{4}}\text{. Now I sub out }a: x=\dfrac{2+1-{{a}^{4}}}{{{a}^{3}}+b}=\dfrac{2+1-{{\left( \dfrac{1}{b} \right)}^{4}}}{{{\left( \dfrac{1}{b} \right)}^{3}}+b}=\dfrac{\dfrac{3{{b}^{4}}-1}{{{b}^{4}}}}{\dfrac{1+{{b}^{3}}}{{{b}^{3}}}}=\dfrac{3b^4 - 1}{1 + b^3}.\text{ This is NOT }-b.$

 

[mighty2000]

Dear forum post author,

Thank you for your question. I see that you have attempted to prove that x=-b using the given equations and points. However, I noticed that in your attempt, you substituted a=1/b, which may have led to the incorrect result. Instead, try substituting a=-1/b, which should give you the desired result of x=-b.

Additionally, when differentiating C1, I believe there may be a mistake in your solution. The correct differentiation of C1 would be d/dx(xy)=y+xdy/dx=0, which leads to dy/dx=-y/x. This may also affect your final result.

I hope this helps. Keep up the good work in your studies!



Scientist