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Question-54447

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Question Number 54447 by ajfour last updated on 03/Feb/19
Commented by ajfour last updated on 14/Feb/19
Find coordinates of A and C.
$${Find}\:{coordinates}\:{of}\:{A}\:{and}\:{C}.\:\:\:\:\:\:\:\: \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19
solve x^2 =ry and (x−r)^2 +(y−r)^2 =r^2 (x−r)^2 +((x^2 /r)−r)^2 =r^2 x^2 −2xr+r^2 +(x^4 /r^2 )−2x^2 +r^2 =r^2 (x^4 /r^2 )−x^2 −2xr+r^2 =0 x^4 −r^2 x^2 −2xr^3 +r^4 =0 on solving we get four value of x so corresponding four value of y thus we can get co iridinate of point A and C ok le t try to solve... let (x/r)=k (x^4 /r^4 )−(x^2 /r^2 )−((2x)/r)+1=0 k^4 −k^2 −2k+1=0 f(k)=k^4 −k^2 −2k+1=k^4 −k^2 −2k−1+2=k^4 +2−(k+1)^2 f(0)>0 f(1)<0 f(0.5)=(1/(16))−(1/4)−(2/2)+1<0 f(0.4)=0.0256+2−1.96>0 so value of [0.5 >k>0.4] thus 0.5>(x/r)>0.4 so 0.5r>x>0.4r thus x=(0.4+ε)r ε=is a small number...
$${solve}\:\:{x}^{\mathrm{2}} ={ry}\:{and}\:\left({x}−{r}\right)^{\mathrm{2}} +\left({y}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\left({x}−{r}\right)^{\mathrm{2}} +\left(\frac{{x}^{\mathrm{2}} }{{r}}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{xr}+{r}^{\mathrm{2}} +\frac{{x}^{\mathrm{4}} }{{r}^{\mathrm{2}} }−\mathrm{2}{x}^{\mathrm{2}} +{r}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\frac{{x}^{\mathrm{4}} }{{r}^{\mathrm{2}} }−{x}^{\mathrm{2}} −\mathrm{2}{xr}+{r}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}^{\mathrm{4}} −{r}^{\mathrm{2}} {x}^{\mathrm{2}} −\mathrm{2}{xr}^{\mathrm{3}} +{r}^{\mathrm{4}} =\mathrm{0} \\ $$$${on}\:{solving}\:\:{we}\:{get}\:{four}\:{value}\:{of}\:{x} \\ $$$${so}\:{corresponding}\:{four}\:{value}\:{of}\:{y} \\ $$$${thus}\:{we}\:{can}\:{get}\:{co}\:{iridinate}\:{of}\:\:{point}\:{A}\:{and}\:{C} \\ $$$${ok}\:{le}\:{t}\:{try}\:{to}\:{solve}… \\ $$$${let}\:\frac{{x}}{{r}}={k} \\ $$$$\frac{{x}^{\mathrm{4}} }{{r}^{\mathrm{4}} }−\frac{{x}^{\mathrm{2}} }{{r}^{\mathrm{2}} }−\frac{\mathrm{2}{x}}{{r}}+\mathrm{1}=\mathrm{0} \\ $$$${k}^{\mathrm{4}} −{k}^{\mathrm{2}} −\mathrm{2}{k}+\mathrm{1}=\mathrm{0} \\ $$$${f}\left({k}\right)={k}^{\mathrm{4}} −{k}^{\mathrm{2}} −\mathrm{2}{k}+\mathrm{1}={k}^{\mathrm{4}} −{k}^{\mathrm{2}} −\mathrm{2}{k}−\mathrm{1}+\mathrm{2}={k}^{\mathrm{4}} +\mathrm{2}−\left({k}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${f}\left(\mathrm{0}\right)>\mathrm{0} \\ $$$${f}\left(\mathrm{1}\right)<\mathrm{0} \\ $$$${f}\left(\mathrm{0}.\mathrm{5}\right)=\frac{\mathrm{1}}{\mathrm{16}}−\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{2}}{\mathrm{2}}+\mathrm{1}<\mathrm{0} \\ $$$${f}\left(\mathrm{0}.\mathrm{4}\right)=\mathrm{0}.\mathrm{0256}+\mathrm{2}−\mathrm{1}.\mathrm{96}>\mathrm{0} \\ $$$${so}\:{value}\:{of}\:\left[\mathrm{0}.\mathrm{5}\:>{k}>\mathrm{0}.\mathrm{4}\right] \\ $$$${thus}\:\:\:\:\mathrm{0}.\mathrm{5}>\frac{{x}}{{r}}>\mathrm{0}.\mathrm{4} \\ $$$${so}\:\:\:\mathrm{0}.\mathrm{5}{r}>{x}>\mathrm{0}.\mathrm{4}{r} \\ $$$${thus}\:\:{x}=\left(\mathrm{0}.\mathrm{4}+\epsilon\right){r}\:\:\:\epsilon={is}\:{a}\:{small}\:{number}… \\ $$
Commented by ajfour last updated on 03/Feb/19
why not solve it, Sir ?
$${why}\:{not}\:{solve}\:{it},\:{Sir}\:? \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19
i can not solve without numirical value.. if numirical value present i can use calculus.. but general eqn...i can not solve...
$${i}\:{can}\:{not}\:{solve}\:{without}\:{numirical}\:{value}.. \\ $$$${if}\:{numirical}\:{value}\:{present}\:{i}\:{can}\:{use}\:{calculus}.. \\ $$$${but}\:{general}\:{eqn}…{i}\:{can}\:{not}\:{solve}… \\ $$
Commented by ajfour last updated on 03/Feb/19
it can be converted to one like numerical value, but still let us not use calculator.
$${it}\:{can}\:{be}\:{converted}\:{to}\:{one}\:{like} \\ $$$${numerical}\:{value},\:{but}\:{still}\:{let}\:{us}\:{not} \\ $$$${use}\:{calculator}. \\ $$
Commented by ajfour last updated on 03/Feb/19
thanks for trying, Sir.
$${thanks}\:{for}\:{trying},\:{Sir}. \\ $$
Answered by ajfour last updated on 04/Feb/19
(x−r)^2 +(y−r)^2 =r^2 (circle eq.) y=x^2 /r (parabola eq.) for x= x_A , x_C x^2 −2rx+(x^4 /r^2 )−2x^2 +r^2 =0 (x^4 /r^2 )−x^2 −2rx+r^2 =0 if x/r =t ⇒ t^4 −t^2 −2t+1=0 ....(i) let us assume equivalently (t^2 −pt+q)(t^2 +pt+(1/q))=0 ⇒ t^4 +(q+(1/q)−p^2 )t^2 +(pq−(p/q))t+1=0 comparing with (i) q+(1/q)= p^2 −1 and q−(1/q)=−(2/p) ⇒ (p^2 −1)^2 −(4/p^2 ) = 4 let us call p^2 =s ⇒ s(s−1)^2 −4s−4=0 let s−1= u ⇒ (u+1)u^2 −4(u+2)=0 ⇒ u^3 +u^2 −4u−8=0 let u=z−1/3 ⇒ z^3 −z^2 _− +(z/3)−(1/(27))+z^2 _− −((2z)/3)+(1/9) −4z+(4/3)−8=0 or z^3 −((13)/3)z−((178)/(27))=0 z= (((89)/(27))+(√(((89^2 )/(27^2 ))−((13^3 )/(27^2 )))) )^(1/3) +(((89)/(27))−(√((((89)/(27^2 ))−((13^3 )/(27^2 )))) )^(1/3) =(1/3){(89+6(√(159)))^(1/3) +(89−6(√(159)))^(1/3) now p^2 =s=u+1 = z+(2/3) since t_A >0 and t_C >0 let′s first choose (t^2 −pt+q)=0 with p_1 >0 p_1 =(√((2+(89+6(√(159)))^(1/3) +(89−6(√(159)))^(1/3) )/3)) ≈ 1.81226 (D/4)= (p^2 /4)−q as q= (((p^2 −1))/2)−(1/p) (D/4)=(1/2)+(1/p)−(p^2 /4) ≈ 0.23073 if it is positive then __________________________ with p_1 =(√((2+(89+6(√(159)))^(1/3) +(89−6(√(159)))^(1/3) )/3)) x_A =r( (p_1 /2)−(√((1/2)+(1/p_1 )−(p_1 ^2 /4)))) ⇒ x_A ≈ 0.4258 r , y_A =x_A ^2 /r x_C =r( (p_1 /2)+(√((1/2)+(1/p_1 )−(p_1 ^2 /4)))) ⇒ x_C ≈ 1.3865 r , y_C =x_C ^2 /r __________________________■
$$\left({x}−{r}\right)^{\mathrm{2}} +\left({y}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \:\:\:\:\left({circle}\:{eq}.\right) \\ $$$${y}={x}^{\mathrm{2}} /{r}\:\:\:\:\:\:\:\:\:\:\:\:\:\left({parabola}\:{eq}.\right) \\ $$$${for}\:{x}=\:{x}_{{A}} ,\:{x}_{{C}} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{rx}+\frac{{x}^{\mathrm{4}} }{{r}^{\mathrm{2}} }−\mathrm{2}{x}^{\mathrm{2}} +{r}^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\frac{{x}^{\mathrm{4}} }{{r}^{\mathrm{2}} }−{x}^{\mathrm{2}} −\mathrm{2}{rx}+{r}^{\mathrm{2}} =\mathrm{0} \\ $$$${if}\:\:\:{x}/{r}\:={t} \\ $$$$\:\Rightarrow\:\:\:{t}^{\mathrm{4}} −{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}=\mathrm{0}\:\:\:\:\:\:….\left({i}\right) \\ $$$${let}\:{us}\:{assume}\:{equivalently} \\ $$$$\:\:\:\:\:\left({t}^{\mathrm{2}} −{pt}+{q}\right)\left({t}^{\mathrm{2}} +{pt}+\frac{\mathrm{1}}{{q}}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{t}^{\mathrm{4}} +\left({q}+\frac{\mathrm{1}}{{q}}−{p}^{\mathrm{2}} \right){t}^{\mathrm{2}} +\left({pq}−\frac{{p}}{{q}}\right){t}+\mathrm{1}=\mathrm{0} \\ $$$${comparing}\:{with}\:\left({i}\right) \\ $$$$\:\:\:\:\boldsymbol{{q}}+\frac{\mathrm{1}}{\boldsymbol{{q}}}=\:\boldsymbol{{p}}^{\mathrm{2}} −\mathrm{1}\:\:\:{and}\:\:\boldsymbol{{q}}−\frac{\mathrm{1}}{\boldsymbol{{q}}}=−\frac{\mathrm{2}}{\boldsymbol{{p}}} \\ $$$$\Rightarrow\:\:\left({p}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} −\frac{\mathrm{4}}{{p}^{\mathrm{2}} }\:=\:\mathrm{4} \\ $$$${let}\:{us}\:{call}\:\:{p}^{\mathrm{2}} ={s} \\ $$$$\Rightarrow\:\:{s}\left({s}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}{s}−\mathrm{4}=\mathrm{0} \\ $$$${let}\:{s}−\mathrm{1}=\:{u} \\ $$$$\Rightarrow\:\:\left({u}+\mathrm{1}\right){u}^{\mathrm{2}} −\mathrm{4}\left({u}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\:{u}^{\mathrm{3}} +{u}^{\mathrm{2}} −\mathrm{4}{u}−\mathrm{8}=\mathrm{0} \\ $$$${let}\:\:{u}={z}−\mathrm{1}/\mathrm{3} \\ $$$$\Rightarrow\:\:{z}^{\mathrm{3}} −\underset{−} {{z}}^{\mathrm{2}} +\frac{{z}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{27}}+\underset{−} {{z}}^{\mathrm{2}} −\frac{\mathrm{2}{z}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{9}} \\ $$$$\:\:\:\:\:\:\:\:\:\:−\mathrm{4}{z}+\frac{\mathrm{4}}{\mathrm{3}}−\mathrm{8}=\mathrm{0} \\ $$$${or}\:\:\:{z}^{\mathrm{3}} −\frac{\mathrm{13}}{\mathrm{3}}{z}−\frac{\mathrm{178}}{\mathrm{27}}=\mathrm{0} \\ $$$$\:\:\:\:{z}=\:\left(\frac{\mathrm{89}}{\mathrm{27}}+\sqrt{\frac{\mathrm{89}^{\mathrm{2}} }{\mathrm{27}^{\mathrm{2}} }−\frac{\mathrm{13}^{\mathrm{3}} }{\mathrm{27}^{\mathrm{2}} }}\:\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left(\frac{\mathrm{89}}{\mathrm{27}}−\sqrt{\left(\frac{\mathrm{89}}{\mathrm{27}^{\mathrm{2}} }−\frac{\mathrm{13}^{\mathrm{3}} }{\mathrm{27}^{\mathrm{2}} }\right.}\:\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{3}}\left\{\left(\mathrm{89}+\mathrm{6}\sqrt{\mathrm{159}}\right)^{\mathrm{1}/\mathrm{3}} +\left(\mathrm{89}−\mathrm{6}\sqrt{\mathrm{159}}\right)^{\mathrm{1}/\mathrm{3}} \right. \\ $$$${now}\:\:\:{p}^{\mathrm{2}} ={s}={u}+\mathrm{1}\:=\:{z}+\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${since}\:{t}_{{A}} >\mathrm{0}\:{and}\:{t}_{{C}} \:>\mathrm{0} \\ $$$${let}'{s}\:{first}\:{choose}\:\:\left({t}^{\mathrm{2}} −{pt}+{q}\right)=\mathrm{0} \\ $$$${with}\:{p}_{\mathrm{1}} >\mathrm{0}\: \\ $$$${p}_{\mathrm{1}} =\sqrt{\frac{\mathrm{2}+\left(\mathrm{89}+\mathrm{6}\sqrt{\mathrm{159}}\right)^{\mathrm{1}/\mathrm{3}} +\left(\mathrm{89}−\mathrm{6}\sqrt{\mathrm{159}}\right)^{\mathrm{1}/\mathrm{3}} }{\mathrm{3}}} \\ $$$$\:\:\:\:\approx\:\mathrm{1}.\mathrm{81226} \\ $$$$\frac{{D}}{\mathrm{4}}=\:\frac{{p}^{\mathrm{2}} }{\mathrm{4}}−{q}\: \\ $$$$\:{as}\:\:{q}=\:\frac{\left({p}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{2}}−\frac{\mathrm{1}}{{p}} \\ $$$$\frac{{D}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{{p}}−\frac{{p}^{\mathrm{2}} }{\mathrm{4}}\:\approx\:\mathrm{0}.\mathrm{23073} \\ $$$${if}\:{it}\:{is}\:{positive}\:{then} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$${with}\: \\ $$$${p}_{\mathrm{1}} =\sqrt{\frac{\mathrm{2}+\left(\mathrm{89}+\mathrm{6}\sqrt{\mathrm{159}}\right)^{\mathrm{1}/\mathrm{3}} +\left(\mathrm{89}−\mathrm{6}\sqrt{\mathrm{159}}\right)^{\mathrm{1}/\mathrm{3}} }{\mathrm{3}}} \\ $$$$\:\:{x}_{{A}} ={r}\left(\:\frac{{p}_{\mathrm{1}} }{\mathrm{2}}−\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{{p}_{\mathrm{1}} }−\frac{{p}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{4}}}\right) \\ $$$$\Rightarrow\:\:{x}_{{A}} \approx\:\mathrm{0}.\mathrm{4258}\:{r}\:\:\:,\:\:\:{y}_{{A}} ={x}_{{A}} ^{\mathrm{2}} /{r} \\ $$$$\:\:\:\:\:\:{x}_{{C}} ={r}\left(\:\frac{{p}_{\mathrm{1}} }{\mathrm{2}}+\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{{p}_{\mathrm{1}} }−\frac{{p}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{4}}}\right) \\ $$$$\Rightarrow\:\:{x}_{{C}} \:\approx\:\mathrm{1}.\mathrm{3865}\:{r}\:\:\:,\:\:\:\:{y}_{{C}} ={x}_{{C}} ^{\mathrm{2}} /{r}\: \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\blacksquare \\ $$

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