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Question Number 223865 by ajfour last updated on 07/Aug/25
Commented by ajfour last updated on 07/Aug/25
Contrary to diagram labelling, lets have a=1, b=b. ∠CAE=π/6. If APC is a straight line, P being a point of tangency, then find b, r, R.
$${Contrary}\:{to}\:{diagram}\:{labelling},\:{lets} \\ $$$${have}\:{a}=\mathrm{1},\:{b}={b}.\:\angle{CAE}=\pi/\mathrm{6}. \\ $$$${If}\:{APC}\:{is}\:{a}\:{straight}\:{line},\:{P}\:{being}\:{a} \\ $$$${point}\:{of}\:{tangency},\:{then}\:{find}\:{b},\:{r},\:{R}. \\ $$
Answered by mr W last updated on 08/Aug/25
(x^2 /a^2 )+(y^2 /b^2 )=1 (x/a^2 )+((yy′)/b^2 )=0 let λ=(a^2 /b^2 ) A(R, 0) P(p, q) (p^2 /a^2 )+(q^2 /b^2 )=1 (p/a^2 )−((q(√3))/b^2 )=0 ⇒p=(√3)λq ((3λ^2 q^2 )/a^2 )+(q^2 /b^2 )=1 ⇒q=(b/( (√(1+3λ)))) (r/(R−r))=sin (π/6)=(1/2) ⇒R=3r p−R=(√3)q (√3)λq−3r=(√3)q ⇒q=(((√3)r)/(λ−1)), p=((3λr)/(λ−1)) (√((p−R)^2 +q^2 ))=R−2r=r (p−3r)^2 +q^2 =r^2 (((3λr)/(λ−1))−3r)^2 +((3r^2 )/((λ−1)^2 ))=r^2 ((12)/((λ−1)^2 ))=1 ⇒λ=2(√3)+1=(a^2 /b^2 ) ⇒b=(a/( (√(2(√3)+1))))≈0.473296 q=(b/( (√(1+3λ))))=(((√3)r)/(λ−1)) ⇒r=((a(λ−1))/( (√(3λ(1+3λ)))))
$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{x}}{{a}^{\mathrm{2}} }+\frac{{yy}'}{{b}^{\mathrm{2}} }=\mathrm{0} \\ $$$${let}\:\lambda=\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} } \\ $$$${A}\left({R},\:\mathrm{0}\right) \\ $$$${P}\left({p},\:{q}\right) \\ $$$$\frac{{p}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{q}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{p}}{{a}^{\mathrm{2}} }−\frac{{q}\sqrt{\mathrm{3}}}{{b}^{\mathrm{2}} }=\mathrm{0}\:\Rightarrow{p}=\sqrt{\mathrm{3}}\lambda{q} \\ $$$$\frac{\mathrm{3}\lambda^{\mathrm{2}} {q}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{q}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\Rightarrow{q}=\frac{{b}}{\:\sqrt{\mathrm{1}+\mathrm{3}\lambda}} \\ $$$$\frac{{r}}{{R}−{r}}=\mathrm{sin}\:\frac{\pi}{\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{R}=\mathrm{3}{r} \\ $$$${p}−{R}=\sqrt{\mathrm{3}}{q} \\ $$$$\sqrt{\mathrm{3}}\lambda{q}−\mathrm{3}{r}=\sqrt{\mathrm{3}}{q} \\ $$$$\Rightarrow{q}=\frac{\sqrt{\mathrm{3}}{r}}{\lambda−\mathrm{1}},\:{p}=\frac{\mathrm{3}\lambda{r}}{\lambda−\mathrm{1}} \\ $$$$\sqrt{\left({p}−{R}\right)^{\mathrm{2}} +{q}^{\mathrm{2}} }={R}−\mathrm{2}{r}={r} \\ $$$$\left({p}−\mathrm{3}{r}\right)^{\mathrm{2}} +{q}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{3}\lambda{r}}{\lambda−\mathrm{1}}−\mathrm{3}{r}\right)^{\mathrm{2}} +\frac{\mathrm{3}{r}^{\mathrm{2}} }{\left(\lambda−\mathrm{1}\right)^{\mathrm{2}} }={r}^{\mathrm{2}} \\ $$$$\frac{\mathrm{12}}{\left(\lambda−\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{1} \\ $$$$\Rightarrow\lambda=\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}=\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} } \\ $$$$\Rightarrow{b}=\frac{{a}}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}}}\approx\mathrm{0}.\mathrm{473296} \\ $$$${q}=\frac{{b}}{\:\sqrt{\mathrm{1}+\mathrm{3}\lambda}}=\frac{\sqrt{\mathrm{3}}{r}}{\lambda−\mathrm{1}} \\ $$$$\Rightarrow{r}=\frac{{a}\left(\lambda−\mathrm{1}\right)}{\:\sqrt{\mathrm{3}\lambda\left(\mathrm{1}+\mathrm{3}\lambda\right)}} \\ $$
Commented by mr W last updated on 08/Aug/25
Commented by mr W last updated on 09/Aug/25
(x^2 /a^2 )+(y^2 /b^2 )=1 (x/a^2 )+((yy′)/b^2 )=0 let t=(1/(tan θ)) (θ=(π/6) as example) let λ=(a^2 /b^2 ) A(R, 0) P(p, q) (p^2 /a^2 )+(q^2 /b^2 )=1 (p/a^2 )−((qt)/b^2 )=0 ⇒p=λtq ((λ^2 t^2 q^2 )/a^2 )+(q^2 /b^2 )=1 ⇒q=(b/( (√(1+λt^2 )))) (r/(R−r))=sin θ=(1/( (√(1+t^2 )))) ⇒R=(1+(√(1+t^2 )))r p−R=tq λtq−(1+(√(1+t^2 )))r=tq ⇒q=(((1+(√(1+t^2 )))r)/((λ−1)t)), p=((λ(1+(√(1+t^2 )))r)/((λ−1))) (√((p−R)^2 +q^2 ))=R−2r=((√(1+t^2 ))−1)r λ=1+((((√(1+t^2 ))+1)^2 (√(1+t^2 )))/t^3 ) r=((a(λ−1)t)/( (1+(√(1+t^2 )))(√(λ(1+λt^2 )))))
$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{x}}{{a}^{\mathrm{2}} }+\frac{{yy}'}{{b}^{\mathrm{2}} }=\mathrm{0} \\ $$$${let}\:{t}=\frac{\mathrm{1}}{\mathrm{tan}\:\theta}\:\:\left(\theta=\frac{\pi}{\mathrm{6}}\:{as}\:{example}\right) \\ $$$${let}\:\lambda=\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} } \\ $$$${A}\left({R},\:\mathrm{0}\right) \\ $$$${P}\left({p},\:{q}\right) \\ $$$$\frac{{p}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{q}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{p}}{{a}^{\mathrm{2}} }−\frac{{qt}}{{b}^{\mathrm{2}} }=\mathrm{0}\:\Rightarrow{p}=\lambda{tq} \\ $$$$\frac{\lambda^{\mathrm{2}} {t}^{\mathrm{2}} {q}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{q}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\Rightarrow{q}=\frac{{b}}{\:\sqrt{\mathrm{1}+\lambda{t}^{\mathrm{2}} }} \\ $$$$\frac{{r}}{{R}−{r}}=\mathrm{sin}\:\theta=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }} \\ $$$$\Rightarrow{R}=\left(\mathrm{1}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right){r} \\ $$$${p}−{R}={tq} \\ $$$$\lambda{tq}−\left(\mathrm{1}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right){r}={tq} \\ $$$$\Rightarrow{q}=\frac{\left(\mathrm{1}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right){r}}{\left(\lambda−\mathrm{1}\right){t}},\:{p}=\frac{\lambda\left(\mathrm{1}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right){r}}{\left(\lambda−\mathrm{1}\right)} \\ $$$$\sqrt{\left({p}−{R}\right)^{\mathrm{2}} +{q}^{\mathrm{2}} }={R}−\mathrm{2}{r}=\left(\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }−\mathrm{1}\right){r} \\ $$$$\lambda=\mathrm{1}+\frac{\left(\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }+\mathrm{1}\right)^{\mathrm{2}} \sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{{t}^{\mathrm{3}} } \\ $$$${r}=\frac{{a}\left(\lambda−\mathrm{1}\right){t}}{\:\left(\mathrm{1}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right)\sqrt{\lambda\left(\mathrm{1}+\lambda{t}^{\mathrm{2}} \right)}} \\ $$
Commented by mr W last updated on 09/Aug/25
Commented by ajfour last updated on 10/Aug/25
thats great! effort and presentation sir.
$${thats}\:{great}!\:{effort}\:{and}\:{presentation} \\ $$$${sir}. \\ $$
Commented by ajfour last updated on 10/Aug/25
https://g.co/gemini/share/604fc367abb6

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