Question-214483

Question Number 214483 by MATHEMATICSAM last updated on 09/Dec/24

Answered by ajfour last updated on 09/Dec/24

let C≡(h, 2−r) (x−h)^2 +(mx^2 −2+r)^2 =r^2 ((h−x)/(mx^2 −2+r))=2mx (mx^2 −2+r)^2 (4m^2 x^2 +1)=r^2 z^2 {4m(z+2−r)+1}=r^2 z^3 +(2−r+(1/(4m)))z^2 =(r^2 /(4m)) (1/z^3 )−((4m)/r^2 )(2+(1/(4m))−r)(1/z)−((4m)/r^2 )=0 D=0 ⇒ f(m)=(4m)(2+(1/(4m))−r)^3 =((27r^2 )/4) f(1)=f((1/4)) ⇒ 4(2+(1/4)−r)^3 =(2+1−r)^3 ⇒ 2^(2/3) ((9/4)−r)=3−r (9/2^(4/3) )−3=r(2^(2/3) −1) r=((9−6×2^(1/3) )/(4−2×2^(1/3) )) ≈ 0.9732 cant be true..

$${let}\:{C}\equiv\left({h},\:\mathrm{2}−{r}\right) \\ $$$$\left({x}−{h}\right)^{\mathrm{2}} +\left({mx}^{\mathrm{2}} −\mathrm{2}+{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\frac{{h}−{x}}{{mx}^{\mathrm{2}} −\mathrm{2}+{r}}=\mathrm{2}{mx} \\ $$$$\left({mx}^{\mathrm{2}} −\mathrm{2}+{r}\right)^{\mathrm{2}} \left(\mathrm{4}{m}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{1}\right)={r}^{\mathrm{2}} \\ $$$${z}^{\mathrm{2}} \left\{\mathrm{4}{m}\left({z}+\mathrm{2}−{r}\right)+\mathrm{1}\right\}={r}^{\mathrm{2}} \\ $$$${z}^{\mathrm{3}} +\left(\mathrm{2}−{r}+\frac{\mathrm{1}}{\mathrm{4}{m}}\right){z}^{\mathrm{2}} =\frac{{r}^{\mathrm{2}} }{\mathrm{4}{m}} \\ $$$$\frac{\mathrm{1}}{{z}^{\mathrm{3}} }−\frac{\mathrm{4}{m}}{{r}^{\mathrm{2}} }\left(\mathrm{2}+\frac{\mathrm{1}}{\mathrm{4}{m}}−{r}\right)\frac{\mathrm{1}}{{z}}−\frac{\mathrm{4}{m}}{{r}^{\mathrm{2}} }=\mathrm{0} \\ $$$${D}=\mathrm{0}\:\:\Rightarrow \\ $$$${f}\left({m}\right)=\left(\mathrm{4}{m}\right)\left(\mathrm{2}+\frac{\mathrm{1}}{\mathrm{4}{m}}−{r}\right)^{\mathrm{3}} =\frac{\mathrm{27}{r}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${f}\left(\mathrm{1}\right)={f}\left(\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$\Rightarrow\:\:\mathrm{4}\left(\mathrm{2}+\frac{\mathrm{1}}{\mathrm{4}}−{r}\right)^{\mathrm{3}} =\left(\mathrm{2}+\mathrm{1}−{r}\right)^{\mathrm{3}} \\ $$$$\Rightarrow\:\:\mathrm{2}^{\mathrm{2}/\mathrm{3}} \left(\frac{\mathrm{9}}{\mathrm{4}}−{r}\right)=\mathrm{3}−{r} \\ $$$$\frac{\mathrm{9}}{\mathrm{2}^{\mathrm{4}/\mathrm{3}} }−\mathrm{3}={r}\left(\mathrm{2}^{\mathrm{2}/\mathrm{3}} −\mathrm{1}\right) \\ $$$${r}=\frac{\mathrm{9}−\mathrm{6}×\mathrm{2}^{\mathrm{1}/\mathrm{3}} }{\mathrm{4}−\mathrm{2}×\mathrm{2}^{\mathrm{1}/\mathrm{3}} }\:\approx\:\mathrm{0}.\mathrm{9732} \\ $$$${cant}\:{be}\:{true}.. \\ $$

Answered by mr W last updated on 10/Dec/24

circle center at (h, 2−r) touching y=x^2 at (p, q) q=p^2 (dy/dx)=2p=tan α h=p+r sin α=p+((2pr)/( (√(1+4p^2 )))) 2−r=p^2 −r cos α=p^2 −(r/( (√(1+4p^2 )))) 2−p^2 =r(1−(1/( (√(1+4p^2 ))))) { ((r=((2−p^2 )/(1−(1/( (√(1+4p^2 )))))))),((h=p+((2p(2−p^2 ))/( (√(1+4p^2 ))−1)))) :} ...(I) touching y=(x^2 /4) at (u, v) v=(u^2 /4) (dy/dx)=(u/2)=tan β h=u−r sin β=u−((ur)/(2(√(1+(u^2 /4))))) 2−r=(u^2 /4)+r cos β=(u^2 /4)+(r/( (√(1+(u^2 /4))))) { ((r=((2−(u^2 /4))/(1+(1/( (√(1+(u^2 /4))))))))),((h=u−((u(2−(u^2 /4)))/(2((√(1+(u^2 /4)))+1))))) :} ...(II) intersection of both parameterised curves is at h≈1.76833 r≈0.50832 ✓

$${circle}\:{center}\:{at}\:\left({h},\:\mathrm{2}−{r}\right) \\ $$$${touching}\:{y}={x}^{\mathrm{2}} \:{at}\:\left({p},\:{q}\right) \\ $$$${q}={p}^{\mathrm{2}} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{2}{p}=\mathrm{tan}\:\alpha \\ $$$${h}={p}+{r}\:\mathrm{sin}\:\alpha={p}+\frac{\mathrm{2}{pr}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }} \\ $$$$\mathrm{2}−{r}={p}^{\mathrm{2}} −{r}\:\mathrm{cos}\:\alpha={p}^{\mathrm{2}} −\frac{{r}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }} \\ $$$$\mathrm{2}−{p}^{\mathrm{2}} ={r}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }}\right) \\ $$$$\begin{cases}{{r}=\frac{\mathrm{2}−{p}^{\mathrm{2}} }{\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }}}}\\{{h}={p}+\frac{\mathrm{2}{p}\left(\mathrm{2}−{p}^{\mathrm{2}} \right)}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }−\mathrm{1}}}\end{cases}\:\:\:\:\:\:…\left({I}\right) \\ $$$${touching}\:{y}=\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\:{at}\:\left({u},\:{v}\right) \\ $$$${v}=\frac{{u}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\frac{{dy}}{{dx}}=\frac{{u}}{\mathrm{2}}=\mathrm{tan}\:\beta \\ $$$${h}={u}−{r}\:\mathrm{sin}\:\beta={u}−\frac{{ur}}{\mathrm{2}\sqrt{\mathrm{1}+\frac{{u}^{\mathrm{2}} }{\mathrm{4}}}} \\ $$$$\mathrm{2}−{r}=\frac{{u}^{\mathrm{2}} }{\mathrm{4}}+{r}\:\mathrm{cos}\:\beta=\frac{{u}^{\mathrm{2}} }{\mathrm{4}}+\frac{{r}}{\:\sqrt{\mathrm{1}+\frac{{u}^{\mathrm{2}} }{\mathrm{4}}}} \\ $$$$\begin{cases}{{r}=\frac{\mathrm{2}−\frac{{u}^{\mathrm{2}} }{\mathrm{4}}}{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\frac{{u}^{\mathrm{2}} }{\mathrm{4}}}}}}\\{{h}={u}−\frac{{u}\left(\mathrm{2}−\frac{{u}^{\mathrm{2}} }{\mathrm{4}}\right)}{\mathrm{2}\left(\sqrt{\mathrm{1}+\frac{{u}^{\mathrm{2}} }{\mathrm{4}}}+\mathrm{1}\right)}}\end{cases}\:\:\:…\left({II}\right) \\ $$$${intersection}\:{of}\:{both}\:{parameterised} \\ $$$${curves}\:{is}\:{at} \\ $$$${h}\approx\mathrm{1}.\mathrm{76833} \\ $$$${r}\approx\mathrm{0}.\mathrm{50832}\:\checkmark \\ $$

Commented by TonyCWX08 last updated on 10/Dec/24

May I know why there is an angle there? I don′t see the α and β...

$${May}\:{I}\:{know}\:{why}\:{there}\:{is}\:{an}\:{angle}\:{there}? \\ $$$${I}\:{don}'{t}\:{see}\:{the}\:\alpha\:{and}\:\beta… \\ $$

Commented by mr W last updated on 10/Dec/24

Commented by mr W last updated on 10/Dec/24

Commented by TonyCWX08 last updated on 11/Dec/24

$${Where}\:{can}\:{I}\:{actually}\:{learn}\:{all}\:{these}\:{stuffs}?? \\ $$

Commented by ajfour last updated on 11/Dec/24

You cant find heaven anywhere other than where you are.��

Commented by mr W last updated on 11/Dec/24

$${To}\:{learn}\:{things}\:{is}\:{one}\:{thing},\:{to} \\ $$$${apply}\:{the}\:{things}\:{learnt}\:{is}\:{an}\:{other} \\ $$$${thing}! \\ $$$${Perhaps}\:{you}\:{have}\:{already}\:{learnt} \\ $$$${all}\:{the}\:{stuff}\:{concerned}\:{in}\:{this}\: \\ $$$${question},\:{you}\:{only}\:{need}\:{to}\:{apply} \\ $$$${it}. \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply