Question Number 88310 by ajfour last updated on 09/Apr/20
Commented by ajfour last updated on 09/Apr/20
$${Q}.\mathrm{88272}\:\left({revisited}\right) \\ $$
Answered by ajfour last updated on 10/Apr/20
![h^2 +k^2 =r^2 y^2 =b^2 (1−(((x−a)^2 )/a^2 )) [ellipse] x^2 +y^2 −2hx−2ky=0 [circle] (x^2 +y^2 −2hx)^2 =4(r^2 −h^2 )y^2 ⇒ {x^2 +b^2 (1−(((x−a)^2 )/a^2 ))−2hx}^2 = 4(r^2 −h^2 )b^2 (1−(((x−a)^2 )/a^2 )) roots of above eq. are x_1 =0 , x_2 =2h, x_3 =x_4 =t first we eliminate x=0, then determine h(r). then eliminate t and eventually find r. .... let (x/a)=X, (h/a)=s, (b^2 /a^2 )=q, (t/a)=m, (r/a)=R . {X^2 +q(2X−X^2 )−2sX}^2 = 4(R^2 −s^2 )q(2X−X^2 ) ⇒ X{X+q(2−X)−2s}^2 =4q(R^2 −s^2 )(2−X) roots are now X=2s, m,m. lets put X=2s ⇒qs(1−s)=(R^2 −s^2 ) ⇒ R^2 =(1−q)s^2 +qs ...(I) And R^2 −s^2 =qs(1−s) ...(II) now generally (1−q)^2 X^3 +4(1−q)(q−s)X^2 +4(q−s)^2 X +4q(R^2 −s^2 )X−8q(R^2 −s^2 )=0 ⇒ (1−q)^2 X^3 +4(1−q)(q−s)X^2 +4{(q−s)^2 +q(R^2 −s^2 )}X −8q(R^2 −s^2 )=0 2m+2s=−((4(q−s))/((1−q))) ...(i) 2m^2 s=((8q(R^2 −s^2 ))/((1−q)^2 )) ...(ii) ⇒ s{((2(s−q))/((1−q)))−s}^2 =((4q(R^2 −s^2 ))/((1−q)^2 )) now using (II) ⇒ s{((2(s−q))/((1−q)))−s}^2 =((4q(qs)(1−s))/((1−q)^2 )) ⇒ {((2(s−q))/((1−q)))−s}^2 =((4q^2 (1−s))/((1−q)^2 )) ⇒{(1+q)s−2q}^2 =4q^2 (1−s) (1+q)^2 s^2 −4qs=0 and as s≠0 s=((4q)/((1+q)^2 )) Now using (I) ⇒ R=(√((1−q)s^2 +qs )) ⇒ R = (√(((16(1−q)q^2 )/((1+q)^4 ))+((4q^2 )/((1+q)^2 )))) or R=((2q(√((1−q)^2 +4)))/((1+q)^2 )) for example if a=4, b=3 q=(9/(16)), R=((18(√(1073)))/(625)) ≈ 0.9434 ...](https://www.tinkutara.com/question/Q88312.png)
$${h}^{\mathrm{2}} +{k}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${y}^{\mathrm{2}} ={b}^{\mathrm{2}} \left(\mathrm{1}−\frac{\left({x}−{a}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)\:\:\:\:\:\:\:\left[{ellipse}\right] \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{hx}−\mathrm{2}{ky}=\mathrm{0}\:\:\:\left[{circle}\right] \\ $$$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{hx}\right)^{\mathrm{2}} =\mathrm{4}\left({r}^{\mathrm{2}} −{h}^{\mathrm{2}} \right){y}^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$$\left\{{x}^{\mathrm{2}} +{b}^{\mathrm{2}} \left(\mathrm{1}−\frac{\left({x}−{a}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)−\mathrm{2}{hx}\right\}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:=\:\mathrm{4}\left({r}^{\mathrm{2}} −{h}^{\mathrm{2}} \right){b}^{\mathrm{2}} \left(\mathrm{1}−\frac{\left({x}−{a}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right) \\ $$$${roots}\:{of}\:{above}\:{eq}.\:{are} \\ $$$${x}_{\mathrm{1}} =\mathrm{0}\:,\:{x}_{\mathrm{2}} =\mathrm{2}{h},\:{x}_{\mathrm{3}} ={x}_{\mathrm{4}} ={t} \\ $$$${first}\:{we}\:{eliminate}\:{x}=\mathrm{0}, \\ $$$${then}\:{determine}\:{h}\left({r}\right). \\ $$$${then}\:{eliminate}\:{t}\:{and} \\ $$$${eventually}\:{find}\:\boldsymbol{{r}}.\: \\ $$$$…. \\ $$$${let}\:\frac{{x}}{{a}}={X},\:\frac{{h}}{{a}}={s},\:\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }={q},\:\:\frac{{t}}{{a}}={m}, \\ $$$$\frac{{r}}{{a}}={R}\:.\: \\ $$$$\left\{{X}^{\mathrm{2}} +{q}\left(\mathrm{2}{X}−{X}^{\mathrm{2}} \right)−\mathrm{2}{sX}\right\}^{\mathrm{2}} \\ $$$$\:\:\:\:\:=\:\mathrm{4}\left({R}^{\mathrm{2}} −{s}^{\mathrm{2}} \right){q}\left(\mathrm{2}{X}−{X}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\:{X}\left\{{X}+{q}\left(\mathrm{2}−{X}\right)−\mathrm{2}{s}\right\}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:=\mathrm{4}{q}\left({R}^{\mathrm{2}} −{s}^{\mathrm{2}} \right)\left(\mathrm{2}−{X}\right) \\ $$$${roots}\:{are}\:{now}\:\:{X}=\mathrm{2}{s},\:{m},{m}. \\ $$$${lets}\:{put}\:{X}=\mathrm{2}{s} \\ $$$$\Rightarrow{qs}\left(\mathrm{1}−{s}\right)=\left({R}^{\mathrm{2}} −{s}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\:\:{R}^{\mathrm{2}} =\left(\mathrm{1}−{q}\right){s}^{\mathrm{2}} +{qs}\:\:\:\:\:\:…\left({I}\right) \\ $$$${And}\:\:{R}^{\mathrm{2}} −{s}^{\mathrm{2}} ={qs}\left(\mathrm{1}−{s}\right)\:\:\:…\left({II}\right) \\ $$$${now}\:{generally} \\ $$$$\left(\mathrm{1}−{q}\right)^{\mathrm{2}} {X}^{\mathrm{3}} +\mathrm{4}\left(\mathrm{1}−{q}\right)\left({q}−{s}\right){X}^{\mathrm{2}} +\mathrm{4}\left({q}−{s}\right)^{\mathrm{2}} {X} \\ $$$$\:\:+\mathrm{4}{q}\left({R}^{\mathrm{2}} −{s}^{\mathrm{2}} \right){X}−\mathrm{8}{q}\left({R}^{\mathrm{2}} −{s}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{1}−{q}\right)^{\mathrm{2}} {X}^{\mathrm{3}} +\mathrm{4}\left(\mathrm{1}−{q}\right)\left({q}−{s}\right){X}^{\mathrm{2}} \\ $$$$\:\:+\mathrm{4}\left\{\left({q}−{s}\right)^{\mathrm{2}} +{q}\left({R}^{\mathrm{2}} −{s}^{\mathrm{2}} \right)\right\}{X} \\ $$$$\:\:−\mathrm{8}{q}\left({R}^{\mathrm{2}} −{s}^{\mathrm{2}} \right)=\mathrm{0}\: \\ $$$$\mathrm{2}{m}+\mathrm{2}{s}=−\frac{\mathrm{4}\left({q}−{s}\right)}{\left(\mathrm{1}−{q}\right)}\:\:\:\:…\left({i}\right) \\ $$$$\mathrm{2}{m}^{\mathrm{2}} {s}=\frac{\mathrm{8}{q}\left({R}^{\mathrm{2}} −{s}^{\mathrm{2}} \right)}{\left(\mathrm{1}−{q}\right)^{\mathrm{2}} }\:\:\:\:…\left({ii}\right) \\ $$$$\Rightarrow\:\:\:{s}\left\{\frac{\mathrm{2}\left({s}−{q}\right)}{\left(\mathrm{1}−{q}\right)}−{s}\right\}^{\mathrm{2}} =\frac{\mathrm{4}{q}\left({R}^{\mathrm{2}} −{s}^{\mathrm{2}} \right)}{\left(\mathrm{1}−{q}\right)^{\mathrm{2}} } \\ $$$${now}\:{using}\:\left({II}\right) \\ $$$$\Rightarrow\:\:\:{s}\left\{\frac{\mathrm{2}\left({s}−{q}\right)}{\left(\mathrm{1}−{q}\right)}−{s}\right\}^{\mathrm{2}} =\frac{\mathrm{4}{q}\left({qs}\right)\left(\mathrm{1}−{s}\right)}{\left(\mathrm{1}−{q}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\:\left\{\frac{\mathrm{2}\left({s}−{q}\right)}{\left(\mathrm{1}−{q}\right)}−{s}\right\}^{\mathrm{2}} =\frac{\mathrm{4}{q}^{\mathrm{2}} \left(\mathrm{1}−{s}\right)}{\left(\mathrm{1}−{q}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\left\{\left(\mathrm{1}+{q}\right){s}−\mathrm{2}{q}\right\}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:=\mathrm{4}{q}^{\mathrm{2}} \left(\mathrm{1}−{s}\right) \\ $$$$\left(\mathrm{1}+{q}\right)^{\mathrm{2}} {s}^{\mathrm{2}} −\mathrm{4}{qs}=\mathrm{0}\:\:\:\: \\ $$$${and}\:{as}\:{s}\neq\mathrm{0} \\ $$$${s}=\frac{\mathrm{4}{q}}{\left(\mathrm{1}+{q}\right)^{\mathrm{2}} } \\ $$$$\:{Now}\:{using}\:\left({I}\right) \\ $$$$\Rightarrow\:\:{R}=\sqrt{\left(\mathrm{1}−{q}\right){s}^{\mathrm{2}} +{qs}\:}\: \\ $$$$\:\Rightarrow\:{R}\:=\:\sqrt{\frac{\mathrm{16}\left(\mathrm{1}−{q}\right){q}^{\mathrm{2}} }{\left(\mathrm{1}+{q}\right)^{\mathrm{4}} }+\frac{\mathrm{4}{q}^{\mathrm{2}} }{\left(\mathrm{1}+{q}\right)^{\mathrm{2}} }} \\ $$$${or}\:\:\:{R}=\frac{\mathrm{2}{q}\sqrt{\left(\mathrm{1}−{q}\right)^{\mathrm{2}} +\mathrm{4}}}{\left(\mathrm{1}+{q}\right)^{\mathrm{2}} } \\ $$$${for}\:{example}\:{if}\:{a}=\mathrm{4},\:{b}=\mathrm{3} \\ $$$${q}=\frac{\mathrm{9}}{\mathrm{16}},\:\:\:{R}=\frac{\mathrm{18}\sqrt{\mathrm{1073}}}{\mathrm{625}}\:\approx\:\mathrm{0}.\mathrm{9434} \\ $$$$… \\ $$
Commented by mr W last updated on 10/Apr/20
$${it}'{s}\:{perfectly}\:{solved}! \\ $$
Commented by ajfour last updated on 10/Apr/20
$${great},\:{thanks}\:{for}\:{the}\:{confirmation}, \\ $$$${Sir},\:{I}'{m}\:{too}\:{glad}\:{i}\:{cud}\:{solve}\:{it}. \\ $$
Commented by mr W last updated on 09/Apr/20
$${that}'{s}\:{right}\:{sir}. \\ $$
Commented by ajfour last updated on 10/Apr/20
$${thanks}\:{to}\:{mjS}\:{Sir}. \\ $$
Commented by ajfour last updated on 10/Apr/20
Commented by MJS last updated on 10/Apr/20
$$\mathrm{I}\:\mathrm{tried}\:\mathrm{the}\:\mathrm{same}\:\mathrm{path}\:\mathrm{I}\:\mathrm{used}\:\mathrm{with}\:\mathrm{the}\:\mathrm{parabola} \\ $$$$\mathrm{and}\:\mathrm{also}\:\mathrm{get}\:{R}=\frac{\mathrm{18}\sqrt{\mathrm{1073}}}{\mathrm{625}}\:\mathrm{for}\:\frac{{a}}{{b}}=\frac{\mathrm{4}}{\mathrm{3}}\:\mathrm{but}\:\mathrm{there} \\ $$$$\mathrm{should}\:\mathrm{be}\:\mathrm{a}\:\mathrm{scaling}\:\mathrm{factor},\:\mathrm{I}\:\mathrm{calculated}\:\mathrm{with} \\ $$$${a}=\mathrm{1}\wedge{b}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$
Commented by ajfour last updated on 10/Apr/20
$${so}\:{our}\:{answer}\:{should}\:{be}\:{right} \\ $$$${then}..?\:\:{required}\:{radius}\:{r}={aR},\:{Sir}. \\ $$
Commented by mr W last updated on 10/Apr/20
$${i}\:{checked}\:{my}\:{result}\:{again}\:{and}\:{found} \\ $$$${an}\:{error}.\:{now}\:{we}\:{have}\:{the}\:{same}\:{result}. \\ $$