Question Number 228378 by ajfour last updated on 10/Apr/26
Commented by ajfour last updated on 10/Apr/26
$${parabola}\:{eq}.\:{is}\:{y}={b}^{\mathrm{2}} −{x}^{\mathrm{2}} \\ $$$${Find}\:{radius}\:{of}\:{circle}\:{in}\:{terms} \\ $$$${of}\:{b}. \\ $$
Answered by mr W last updated on 11/Apr/26
Commented by fantastic2 last updated on 12/Apr/26
$${the}\:{last}\:{line}\:{can}\:{be}\:{simplified}\:{to} \\ $$$$\Rightarrow{h}−\frac{{p}^{\mathrm{2}} }{{k}}=\frac{{p}}{\:\sqrt{\mathrm{4}{p}^{\mathrm{2}} +{k}^{\mathrm{2}} }+\mathrm{2}{p}}\left({k}+\sqrt{\mathrm{4}{p}^{\mathrm{2}} +{k}^{\mathrm{2}} }\right)\:\:\:…\left({ii}\right) \\ $$$$ \\ $$
Commented by mr W last updated on 11/Apr/26
Commented by mr W last updated on 11/Apr/26
$${y}={h}−\frac{{x}^{\mathrm{2}} }{{k}}\:{with}\:{h}={b}^{\mathrm{2}} ,\:{k}=\mathrm{1} \\ $$$${C}\left({r},\:{r}\right) \\ $$$${P}\left({p},\:{h}−\frac{{p}^{\mathrm{2}} }{{k}}\right) \\ $$$$\mathrm{tan}\:\theta=\frac{\mathrm{2}{p}}{{k}}\:\Rightarrow\mathrm{sin}\:\theta=\frac{\mathrm{2}{p}}{\:\sqrt{\mathrm{4}{p}^{\mathrm{2}} +{k}^{\mathrm{2}} }} \\ $$$${p}={r}\left(\mathrm{1}+\mathrm{sin}\:\theta\right)={r}\left(\mathrm{1}+\frac{\mathrm{2}{p}}{\:\sqrt{\mathrm{4}{p}^{\mathrm{2}} +{k}^{\mathrm{2}} }}\right) \\ $$$$\Rightarrow{r}=\frac{{p}}{\mathrm{1}+\frac{\mathrm{2}{p}}{\:\sqrt{\mathrm{4}{p}^{\mathrm{2}} +{k}^{\mathrm{2}} }}}\:\:\:…\left({i}\right) \\ $$$${h}−\frac{{p}^{\mathrm{2}} }{{k}}={r}\left(\mathrm{1}+\mathrm{cos}\:\theta\right) \\ $$$${h}−\frac{{p}^{\mathrm{2}} }{{k}}={r}\left(\mathrm{1}+\frac{{k}}{\:\sqrt{\mathrm{4}{p}^{\mathrm{2}} +{k}^{\mathrm{2}} }}\right) \\ $$$$\Rightarrow{h}−\frac{{p}^{\mathrm{2}} }{{k}}=\frac{{p}}{\mathrm{1}+\frac{\mathrm{2}{p}}{\:\sqrt{\mathrm{4}{p}^{\mathrm{2}} +{k}^{\mathrm{2}} }}}\left(\mathrm{1}+\frac{{k}}{\:\sqrt{\mathrm{4}{p}^{\mathrm{2}} +{k}^{\mathrm{2}} }}\right)\:\:\:…\left({ii}\right) \\ $$$${example}:\: \\ $$$${h}={b}^{\mathrm{2}} =\mathrm{8},\:{k}=\mathrm{2}\:\Rightarrow{r}\approx\mathrm{1}.\mathrm{7349} \\ $$$${h}={b}^{\mathrm{2}} =\mathrm{1},\:{k}=\mathrm{1}\:\Rightarrow{r}\approx\mathrm{0}.\mathrm{3613} \\ $$
Commented by fantastic2 last updated on 11/Apr/26
$${wow}\:{sir}. \\ $$
Commented by ajfour last updated on 15/Apr/26
$${thank}\:{you} \\ $$
Commented by fantastic2 last updated on 15/Apr/26
$${please}\:{post}\:{some}\:{more}\:\:{geometry}\:{q}\:{sir}. \\ $$
Answered by fantastic2 last updated on 11/Apr/26
Commented by Ghisom_ last updated on 11/Apr/26
![@fantastic, I think you already know this: tan α =t ⇔ ((sin α)/(cos α))=t ⇔ ((sin α)/( (√(1−sin^2 α))))=t ⇔ ⇔ ((sin^2 α)/(1−sin^2 α))=t^2 ⇔ sin^2 α =(t^2 /(t^2 +1)) ⇔ sin α =(t/( (√(t^2 +1)))) [similar cos α =(1/( (√(t^2 +1))))] tan α =2p ⇒ sin α =((2p)/( (√(4p^2 +1))))](https://www.tinkutara.com/question/Q228407.png)
$$@\mathrm{fantastic},\:\mathrm{I}\:\mathrm{think}\:\mathrm{you}\:\mathrm{already}\:\mathrm{know}\:\mathrm{this}: \\ $$$$\mathrm{tan}\:\alpha\:={t}\:\Leftrightarrow\:\frac{\mathrm{sin}\:\alpha}{\mathrm{cos}\:\alpha}={t}\:\Leftrightarrow\:\frac{\mathrm{sin}\:\alpha}{\:\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\alpha}}={t}\:\Leftrightarrow \\ $$$$\Leftrightarrow\:\frac{\mathrm{sin}^{\mathrm{2}} \:\alpha}{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\alpha}={t}^{\mathrm{2}} \:\Leftrightarrow\:\mathrm{sin}^{\mathrm{2}} \:\alpha\:=\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} +\mathrm{1}}\:\Leftrightarrow \\ $$$$\mathrm{sin}\:\alpha\:=\frac{{t}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\left[\mathrm{similar}\:\mathrm{cos}\:\alpha\:=\frac{\mathrm{1}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}\right] \\ $$$$ \\ $$$$\mathrm{tan}\:\alpha\:=\mathrm{2}{p}\:\Rightarrow\:\mathrm{sin}\:\alpha\:=\frac{\mathrm{2}{p}}{\:\sqrt{\mathrm{4}{p}^{\mathrm{2}} +\mathrm{1}}} \\ $$
Commented by fantastic2 last updated on 11/Apr/26
$${i}\:{also}\:{got}\:{that}\:\left({see}\:\mathrm{10}{th}\:{line}\right) \\ $$
Commented by fantastic2 last updated on 11/Apr/26
Commented by fantastic2 last updated on 11/Apr/26
![P(p,b^2 −p^2 ) y=b^2 −x^2 m=−2x eq of MN (y−(b^2 −p^2 ))=−2p(x−p) ⇒y=−2px+b^2 +p^2 at x=0 ,y=b^2 +p^2 at y=0 ,x=((b^2 +p^2 )/(2p)) now p=(1+sin α)r p=(1+((2p)/( (√(4p^2 +1)))))r[∵tan α=2p] p=((r((√(4p^2 +1))+2p))/( (√(4p^2 +1)))) △MON=(1/(4p))×(b^2 +p^2 )^2 △MON=(1/2)×r(b^2 +p^2 +((b^2 +p^2 )/(2p))+(√((b^2 +p^2 )^2 +(((b^2 +p^2 )/(2p)))^2 ))) =(r/2)(b^2 +p^2 )(1+(1/(2p))+((√(4p^2 +1))/(2p))) =(r/(4p))(b^2 +p^2 )(2p+1+(√(4p^2 +1))) ⇒(1/(4p))×(b^2 +p^2 )^2 =(r/(4p))(b^2 +p^2 )(2p+1+(√(4p^2 +1))) ⇒b^2 +p^2 =r(2p+1+(√(4p^2 +1))) ⇒b^2 +p^2 =((p(√(4p^2 +1)))/(((√(4p^2 +1))+2p)))((√(4p^2 +1))+2p+1) exaple b=1 p≈0.6466 r≈0.36102](https://www.tinkutara.com/question/Q228403.png)
$${P}\left({p},{b}^{\mathrm{2}} −{p}^{\mathrm{2}} \right) \\ $$$${y}={b}^{\mathrm{2}} −{x}^{\mathrm{2}} \\ $$$${m}=−\mathrm{2}{x} \\ $$$${eq}\:{of}\:{MN} \\ $$$$\left({y}−\left({b}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)\right)=−\mathrm{2}{p}\left({x}−{p}\right) \\ $$$$\Rightarrow{y}=−\mathrm{2}{px}+{b}^{\mathrm{2}} +{p}^{\mathrm{2}} \\ $$$${at}\:{x}=\mathrm{0}\:,{y}={b}^{\mathrm{2}} +{p}^{\mathrm{2}} \\ $$$${at}\:{y}=\mathrm{0}\:,{x}=\frac{{b}^{\mathrm{2}} +{p}^{\mathrm{2}} }{\mathrm{2}{p}} \\ $$$${now}\:{p}=\left(\mathrm{1}+\mathrm{sin}\:\alpha\right){r} \\ $$$${p}=\left(\mathrm{1}+\frac{\mathrm{2}{p}}{\:\sqrt{\mathrm{4}{p}^{\mathrm{2}} +\mathrm{1}}}\right){r}\left[\because\mathrm{tan}\:\alpha=\mathrm{2}{p}\right] \\ $$$${p}=\frac{{r}\left(\sqrt{\mathrm{4}{p}^{\mathrm{2}} +\mathrm{1}}+\mathrm{2}{p}\right)}{\:\sqrt{\mathrm{4}{p}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\bigtriangleup{MON}=\frac{\mathrm{1}}{\mathrm{4}{p}}×\left({b}^{\mathrm{2}} +{p}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\bigtriangleup{MON}=\frac{\mathrm{1}}{\mathrm{2}}×{r}\left({b}^{\mathrm{2}} +{p}^{\mathrm{2}} +\frac{{b}^{\mathrm{2}} +{p}^{\mathrm{2}} }{\mathrm{2}{p}}+\sqrt{\left({b}^{\mathrm{2}} +{p}^{\mathrm{2}} \right)^{\mathrm{2}} +\left(\frac{{b}^{\mathrm{2}} +{p}^{\mathrm{2}} }{\mathrm{2}{p}}\right)^{\mathrm{2}} }\right) \\ $$$$=\frac{{r}}{\mathrm{2}}\left({b}^{\mathrm{2}} +{p}^{\mathrm{2}} \right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{p}}+\frac{\sqrt{\mathrm{4}{p}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{2}{p}}\right) \\ $$$$=\frac{{r}}{\mathrm{4}{p}}\left({b}^{\mathrm{2}} +{p}^{\mathrm{2}} \right)\left(\mathrm{2}{p}+\mathrm{1}+\sqrt{\mathrm{4}{p}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{4}{p}}×\left({b}^{\mathrm{2}} +{p}^{\mathrm{2}} \right)^{\mathrm{2}} =\frac{{r}}{\mathrm{4}{p}}\left({b}^{\mathrm{2}} +{p}^{\mathrm{2}} \right)\left(\mathrm{2}{p}+\mathrm{1}+\sqrt{\mathrm{4}{p}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$$\Rightarrow{b}^{\mathrm{2}} +{p}^{\mathrm{2}} ={r}\left(\mathrm{2}{p}+\mathrm{1}+\sqrt{\mathrm{4}{p}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$$\Rightarrow{b}^{\mathrm{2}} +{p}^{\mathrm{2}} =\frac{{p}\sqrt{\mathrm{4}{p}^{\mathrm{2}} +\mathrm{1}}}{\left(\sqrt{\mathrm{4}{p}^{\mathrm{2}} +\mathrm{1}}+\mathrm{2}{p}\right)}\left(\sqrt{\mathrm{4}{p}^{\mathrm{2}} +\mathrm{1}}+\mathrm{2}{p}+\mathrm{1}\right) \\ $$$${exaple}\:{b}=\mathrm{1} \\ $$$${p}\approx\mathrm{0}.\mathrm{6466} \\ $$$${r}\approx\mathrm{0}.\mathrm{36102} \\ $$
Commented by TonyCWX last updated on 11/Apr/26
$$\boldsymbol{\mathrm{How}}\:\boldsymbol{\mathrm{did}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{get}}\:\boldsymbol{\mathrm{m}}_{\boldsymbol{\mathrm{MN}}} \:=\:−\mathrm{2}? \\ $$