Question-228474

Question Number 228474 by ajfour last updated on 18/Apr/26

Commented by ajfour last updated on 18/Apr/26

Parabola is y=(h−x)^2 Find circle radius in terms of h. How much is r for h=1?

$${Parabola}\:{is}\:\:{y}=\left({h}−{x}\right)^{\mathrm{2}} \\ $$$${Find}\:{circle}\:{radius}\:{in}\:{terms}\:{of}\:{h}. \\ $$$${How}\:{much}\:{is}\:{r}\:{for}\:{h}=\mathrm{1}? \\ $$

Commented by fantastic2 last updated on 18/Apr/26

$${shouldnt}\:{h}\:{be}\:\left({h},\mathrm{0}\right)\:{not}\:\left(\mathrm{0},{h}\right) \\ $$

Answered by TonyCWX last updated on 18/Apr/26

x^2 + y^2 = r^2 y = (h−x)^2 ⇒ y′ = −2(h−p) m_(OP) = (((h−p)^2 )/p) ⇒ m_⊥ = −(p/((h−p)^2 )) −2(h−p) = −(p/((h−p)^2 )) 2(h−p)^3 = p 2(h^3 −3h^2 p+3hp^2 −p^3 )=p −2p^3 +6hp^2 −(6h^2 +1)p+2h^3 = 0 p^3 − 3hp^2 + (3h^2 +(1/2))p − h^3 = 0 p = λ + h ⇒ λ^3 + (1/2)λ + (h/2) = 0 ⇒ λ = ((−(h/4)+(√((27h^2 +2)/(432)))))^(1/3) + ((−(h/4) − (√((27h^2 +2)/(432)))))^(1/3) ⇒ p =h + ((−(h/4)+(√((27h^2 +2)/(432)))))^(1/3) + ((−(h/4) − (√((27h^2 +2)/(432)))))^(1/3) r = (√((h−p)^4 − p^2 )) r = (√((((−(h/4)+(√((27h^2 + 2)/(432)))))^(1/3) + ((−(h/4)−(√((27h^2 + 2)/(432)))))^(1/3) )^4 − (h+((−(h/4)+(√((27h^2 + 2)/(432)))))^(1/3) + ((−(h/4)−(√((27h^2 + 2)/(432)))))^(1/3) )^2 )) h = 1, p = 1+((−(1/4)+(√((29)/(432)))))^(1/3) + ((−(1/4)−(√((29)/(432)))))^(1/3) r = (√((−((−(1/4)+(√((29)/(432)))))^(1/3) −(√(−(1/4)−(√((29)/(432))))))^4 − (1 + ((−(1/4)+(√((29)/(432)))))^(1/3) + (√(−(1/4)−(√((29)/(432))))))^2 )) ≈ 0.537841448698...

$${x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:=\:{r}^{\mathrm{2}} \\ $$$${y}\:=\:\left({h}−{x}\right)^{\mathrm{2}} \:\Rightarrow\:{y}'\:=\:−\mathrm{2}\left({h}−{p}\right) \\ $$$$ \\ $$$${m}_{{OP}} \:=\:\frac{\left({h}−{p}\right)^{\mathrm{2}} }{{p}}\:\Rightarrow\:{m}_{\bot} \:=\:−\frac{{p}}{\left({h}−{p}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$−\mathrm{2}\left({h}−{p}\right)\:=\:−\frac{{p}}{\left({h}−{p}\right)^{\mathrm{2}} } \\ $$$$\mathrm{2}\left({h}−{p}\right)^{\mathrm{3}} \:=\:{p} \\ $$$$\mathrm{2}\left({h}^{\mathrm{3}} −\mathrm{3}{h}^{\mathrm{2}} {p}+\mathrm{3}{hp}^{\mathrm{2}} −{p}^{\mathrm{3}} \right)={p} \\ $$$$−\mathrm{2}{p}^{\mathrm{3}} +\mathrm{6}{hp}^{\mathrm{2}} −\left(\mathrm{6}{h}^{\mathrm{2}} +\mathrm{1}\right){p}+\mathrm{2}{h}^{\mathrm{3}} \:=\:\mathrm{0} \\ $$$${p}^{\mathrm{3}} \:−\:\mathrm{3}{hp}^{\mathrm{2}} \:+\:\left(\mathrm{3}{h}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\right){p}\:−\:{h}^{\mathrm{3}} \:=\:\mathrm{0} \\ $$$${p}\:=\:\lambda\:+\:{h}\:\Rightarrow\:\lambda^{\mathrm{3}} \:+\:\frac{\mathrm{1}}{\mathrm{2}}\lambda\:+\:\frac{{h}}{\mathrm{2}}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\lambda\:=\:\sqrt[{\mathrm{3}}]{−\frac{{h}}{\mathrm{4}}+\sqrt{\frac{\mathrm{27}{h}^{\mathrm{2}} +\mathrm{2}}{\mathrm{432}}}}\:+\:\sqrt[{\mathrm{3}}]{−\frac{{h}}{\mathrm{4}}\:−\:\sqrt{\frac{\mathrm{27}{h}^{\mathrm{2}} +\mathrm{2}}{\mathrm{432}}}} \\ $$$$\Rightarrow\:{p}\:={h}\:+\:\sqrt[{\mathrm{3}}]{−\frac{{h}}{\mathrm{4}}+\sqrt{\frac{\mathrm{27}{h}^{\mathrm{2}} +\mathrm{2}}{\mathrm{432}}}}\:+\:\sqrt[{\mathrm{3}}]{−\frac{{h}}{\mathrm{4}}\:−\:\sqrt{\frac{\mathrm{27}{h}^{\mathrm{2}} +\mathrm{2}}{\mathrm{432}}}} \\ $$$$ \\ $$$${r}\:=\:\sqrt{\left({h}−{p}\right)^{\mathrm{4}} \:−\:{p}^{\mathrm{2}} } \\ $$$${r}\:=\:\sqrt{\left(\sqrt[{\mathrm{3}}]{−\frac{{h}}{\mathrm{4}}+\sqrt{\frac{\mathrm{27}{h}^{\mathrm{2}} \:+\:\mathrm{2}}{\mathrm{432}}}}\:+\:\sqrt[{\mathrm{3}}]{−\frac{{h}}{\mathrm{4}}−\sqrt{\frac{\mathrm{27}{h}^{\mathrm{2}} \:+\:\mathrm{2}}{\mathrm{432}}}}\right)^{\mathrm{4}} \:−\:\left({h}+\sqrt[{\mathrm{3}}]{−\frac{{h}}{\mathrm{4}}+\sqrt{\frac{\mathrm{27}{h}^{\mathrm{2}} \:+\:\mathrm{2}}{\mathrm{432}}}}\:+\:\sqrt[{\mathrm{3}}]{−\frac{{h}}{\mathrm{4}}−\sqrt{\frac{\mathrm{27}{h}^{\mathrm{2}} \:+\:\mathrm{2}}{\mathrm{432}}}}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$${h}\:=\:\mathrm{1},\:{p}\:=\:\mathrm{1}+\sqrt[{\mathrm{3}}]{−\frac{\mathrm{1}}{\mathrm{4}}+\sqrt{\frac{\mathrm{29}}{\mathrm{432}}}}\:+\:\sqrt[{\mathrm{3}}]{−\frac{\mathrm{1}}{\mathrm{4}}−\sqrt{\frac{\mathrm{29}}{\mathrm{432}}}} \\ $$$${r}\:=\:\sqrt{\left(−\sqrt[{\mathrm{3}}]{−\frac{\mathrm{1}}{\mathrm{4}}+\sqrt{\frac{\mathrm{29}}{\mathrm{432}}}}\:−\sqrt{−\frac{\mathrm{1}}{\mathrm{4}}−\sqrt{\frac{\mathrm{29}}{\mathrm{432}}}}\right)^{\mathrm{4}} \:−\:\left(\mathrm{1}\:+\:\sqrt[{\mathrm{3}}]{−\frac{\mathrm{1}}{\mathrm{4}}+\sqrt{\frac{\mathrm{29}}{\mathrm{432}}}}\:+\:\sqrt{−\frac{\mathrm{1}}{\mathrm{4}}−\sqrt{\frac{\mathrm{29}}{\mathrm{432}}}}\right)^{\mathrm{2}} }\:\approx\:\mathrm{0}.\mathrm{537841448698}… \\ $$

Commented by TonyCWX last updated on 18/Apr/26

Thanks for your compliment! It surely made my day as well.

$$\boldsymbol{\mathrm{Thanks}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{your}}\:\boldsymbol{\mathrm{compliment}}! \\ $$$$\boldsymbol{\mathrm{It}}\:\boldsymbol{\mathrm{surely}}\:\boldsymbol{\mathrm{made}}\:\boldsymbol{\mathrm{my}}\:\boldsymbol{\mathrm{day}}\:\boldsymbol{\mathrm{as}}\:\boldsymbol{\mathrm{well}}.\: \\ $$

Commented by ajfour last updated on 18/Apr/26

$${Thank}\:{you},\:{its}\:{so}\:{perfect}! \\ $$$${correct}\:{answer}. \\ $$

Answered by Ghisom_ last updated on 18/Apr/26

parabola: y=(x−h)^2 circle: x^2 +y^2 =r^2 inserting x^4 −4hx^3 +(6h^2 +1)x^2 −4h^3 x+h^4 −r^2 =0 x=t+h t^4 +t^2 +2ht+h^2 −r^2 =0 we need exactly one double real solution plus 2 conjugated complex solutions. ⇒ for t^4 +Pt^2 +Qt+R=0 we must have 16P^4 R−4P^3 Q^2 −128P^2 R^2 +144PQ^2 R−27Q^4 +256R^3 =0 ⇒ r^6 −((6h^2 −1)/2)r^4 +((48h^4 +20h^2 +1)/(16))r^2 −((h^4 (16h^2 +1))/(16))=0 with r=+(√(u+((6h^2 −1)/6))) we get u^3 +((108h^2 −1)/(48))u+((1458h^4 −270h^2 −1)/(864))=0 which has one real solution for h>0 ⇒ we can solve exactly (but it might not be useful) for h=1 we get u^3 +((107)/(48))u+((1187)/(864))=0 u=(1/(12))((−1187+174(√(87)))^(1/3) −(1187+174(√(87)))^(1/3) ) u≈−.544059909396 ⇒ r≈.537841448698 we get a nice solution for h=3: u=−((23)/6) ⇒ r=(√5)

$$\mathrm{parabola}:\:{y}=\left({x}−{h}\right)^{\mathrm{2}} \\ $$$$\mathrm{circle}:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\mathrm{inserting} \\ $$$${x}^{\mathrm{4}} −\mathrm{4}{hx}^{\mathrm{3}} +\left(\mathrm{6}{h}^{\mathrm{2}} +\mathrm{1}\right){x}^{\mathrm{2}} −\mathrm{4}{h}^{\mathrm{3}} {x}+{h}^{\mathrm{4}} −{r}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}={t}+{h} \\ $$$${t}^{\mathrm{4}} +{t}^{\mathrm{2}} +\mathrm{2}{ht}+{h}^{\mathrm{2}} −{r}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{need}\:\mathrm{exactly}\:\mathrm{one}\:\mathrm{double}\:\mathrm{real}\:\mathrm{solution} \\ $$$$\mathrm{plus}\:\mathrm{2}\:\mathrm{conjugated}\:\mathrm{complex}\:\mathrm{solutions}. \\ $$$$\Rightarrow \\ $$$$\mathrm{for}\:{t}^{\mathrm{4}} +{Pt}^{\mathrm{2}} +\mathrm{Q}{t}+{R}=\mathrm{0}\:\mathrm{we}\:\mathrm{must}\:\mathrm{have} \\ $$$$\mathrm{16}{P}^{\mathrm{4}} {R}−\mathrm{4}{P}^{\mathrm{3}} {Q}^{\mathrm{2}} −\mathrm{128}{P}^{\mathrm{2}} {R}^{\mathrm{2}} +\mathrm{144}{PQ}^{\mathrm{2}} {R}−\mathrm{27}{Q}^{\mathrm{4}} +\mathrm{256}{R}^{\mathrm{3}} =\mathrm{0} \\ $$$$\Rightarrow \\ $$$${r}^{\mathrm{6}} −\frac{\mathrm{6}{h}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}}{r}^{\mathrm{4}} +\frac{\mathrm{48}{h}^{\mathrm{4}} +\mathrm{20}{h}^{\mathrm{2}} +\mathrm{1}}{\mathrm{16}}{r}^{\mathrm{2}} −\frac{{h}^{\mathrm{4}} \left(\mathrm{16}{h}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{16}}=\mathrm{0} \\ $$$$\mathrm{with}\:{r}=+\sqrt{{u}+\frac{\mathrm{6}{h}^{\mathrm{2}} −\mathrm{1}}{\mathrm{6}}}\:\mathrm{we}\:\mathrm{get} \\ $$$${u}^{\mathrm{3}} +\frac{\mathrm{108}{h}^{\mathrm{2}} −\mathrm{1}}{\mathrm{48}}{u}+\frac{\mathrm{1458}{h}^{\mathrm{4}} −\mathrm{270}{h}^{\mathrm{2}} −\mathrm{1}}{\mathrm{864}}=\mathrm{0} \\ $$$$\mathrm{which}\:\mathrm{has}\:\mathrm{one}\:\mathrm{real}\:\mathrm{solution}\:\mathrm{for}\:{h}>\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{we}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{exactly}\:\left(\mathrm{but}\:\mathrm{it}\:\mathrm{might}\:\mathrm{not}\right. \\ $$$$\left.\mathrm{be}\:\mathrm{useful}\right) \\ $$$$ \\ $$$$\mathrm{for}\:{h}=\mathrm{1}\:\mathrm{we}\:\mathrm{get} \\ $$$${u}^{\mathrm{3}} +\frac{\mathrm{107}}{\mathrm{48}}{u}+\frac{\mathrm{1187}}{\mathrm{864}}=\mathrm{0} \\ $$$${u}=\frac{\mathrm{1}}{\mathrm{12}}\left(\left(−\mathrm{1187}+\mathrm{174}\sqrt{\mathrm{87}}\right)^{\mathrm{1}/\mathrm{3}} −\left(\mathrm{1187}+\mathrm{174}\sqrt{\mathrm{87}}\right)^{\mathrm{1}/\mathrm{3}} \right) \\ $$$${u}\approx−.\mathrm{544059909396} \\ $$$$\Rightarrow \\ $$$${r}\approx.\mathrm{537841448698} \\ $$$$ \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{a}\:\mathrm{nice}\:\mathrm{solution}\:\mathrm{for}\:{h}=\mathrm{3}: \\ $$$${u}=−\frac{\mathrm{23}}{\mathrm{6}}\:\Rightarrow\:{r}=\sqrt{\mathrm{5}} \\ $$

Commented by TonyCWX last updated on 18/Apr/26

$$\boldsymbol{\mathrm{This}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{neat}}!! \\ $$

Commented by ajfour last updated on 18/Apr/26

$${Wow}\:{thanks}\:{for}\:{the}\:{special}\:{case} \\ $$$${too}. \\ $$

Commented by Ghisom_ last updated on 18/Apr/26

there is a post somewhere in here (but I could not find it now) containing all the possible solutions of x^4 +px^2 +qx+r=0 depending on p, q, r but I found it in a pdf somewhere, might post it again...

$$\mathrm{there}\:\mathrm{is}\:\mathrm{a}\:\mathrm{post}\:\mathrm{somewhere}\:\mathrm{in}\:\mathrm{here}\:\left(\mathrm{but}\:\mathrm{I}\right. \\ $$$$\left.\mathrm{could}\:\mathrm{not}\:\mathrm{find}\:\mathrm{it}\:\mathrm{now}\right)\:\mathrm{containing}\:\mathrm{all}\:\mathrm{the} \\ $$$$\mathrm{possible}\:\mathrm{solutions}\:\mathrm{of} \\ $$$${x}^{\mathrm{4}} +{px}^{\mathrm{2}} +{qx}+{r}=\mathrm{0} \\ $$$$\mathrm{depending}\:\mathrm{on}\:{p},\:{q},\:{r} \\ $$$$\mathrm{but}\:\mathrm{I}\:\mathrm{found}\:\mathrm{it}\:\mathrm{in}\:\mathrm{a}\:\mathrm{pdf}\:\mathrm{somewhere},\:\mathrm{might} \\ $$$$\mathrm{post}\:\mathrm{it}\:\mathrm{again}… \\ $$

Commented by ajfour last updated on 18/Apr/26

https://youtu.be/WMn-9Dg2WiQ?si=jZ7Mt0oeFJv_q2oi

Answered by fantastic2 last updated on 18/Apr/26

Commented by fantastic2 last updated on 18/Apr/26

y=(h−x)^2 ⇒tan θ=2(h−p) now rsin θ=p &rcos θ=(h−p)^2 ⇒tan θ=(p/((h−p)^2 )) (p/((h−p)^2 ))=2(h−p)⇒(p/2)=(h−p)^3 ⇒h=p+((p/2))^(1/3) r=(p/(sin θ))=(p/((2(h−p))/( (√(1+4(h−p)^2 )))))=((p(√(1+4(h−p)^2 )))/(2(h−p))) ⇒r=((p(√(1+4(p+((p/2))^(1/3) −p)^2 )))/(2(p+((p/2))^(1/3) −p))) r in terms of p at h=1 p≈0.410245487[x coordinate where they touch] r≈0.5378414486982

$${y}=\left({h}−{x}\right)^{\mathrm{2}} \Rightarrow\mathrm{tan}\:\theta=\mathrm{2}\left({h}−{p}\right) \\ $$$${now} \\ $$$${r}\mathrm{sin}\:\theta={p}\:\&{r}\mathrm{cos}\:\theta=\left({h}−{p}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\frac{{p}}{\left({h}−{p}\right)^{\mathrm{2}} } \\ $$$$\frac{{p}}{\left({h}−{p}\right)^{\mathrm{2}} }=\mathrm{2}\left({h}−{p}\right)\Rightarrow\frac{{p}}{\mathrm{2}}=\left({h}−{p}\right)^{\mathrm{3}} \Rightarrow{h}={p}+\sqrt[{\mathrm{3}}]{\frac{{p}}{\mathrm{2}}} \\ $$$${r}=\frac{{p}}{\mathrm{sin}\:\theta}=\frac{{p}}{\frac{\mathrm{2}\left({h}−{p}\right)}{\:\sqrt{\mathrm{1}+\mathrm{4}\left({h}−{p}\right)^{\mathrm{2}} }}}=\frac{{p}\sqrt{\mathrm{1}+\mathrm{4}\left({h}−{p}\right)^{\mathrm{2}} }}{\mathrm{2}\left({h}−{p}\right)} \\ $$$$\Rightarrow{r}=\frac{{p}\sqrt{\mathrm{1}+\mathrm{4}\left({p}+\sqrt[{\mathrm{3}}]{\frac{{p}}{\mathrm{2}}}−{p}\right)^{\mathrm{2}} }}{\mathrm{2}\left({p}+\sqrt[{\mathrm{3}}]{\frac{{p}}{\mathrm{2}}}−{p}\right)} \\ $$$${r}\:{in}\:{terms}\:{of}\:{p} \\ $$$${at}\:{h}=\mathrm{1}\:{p}\approx\mathrm{0}.\mathrm{410245487}\left[{x}\:{coordinate}\:{where}\:{they}\:{touch}\right] \\ $$$${r}\approx\mathrm{0}.\mathrm{5378414486982} \\ $$

Commented by ajfour last updated on 18/Apr/26