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Question Number 197730 by ajfour last updated on 27/Sep/23

Commented by ajfour last updated on 27/Sep/23

Determine r in terms of a,b.

$${Determine}\:{r}\:{in}\:{terms}\:{of}\:{a},{b}. \\ $$

Commented by ajfour last updated on 27/Sep/23

https://youtu.be/0sSIzDZtOBQ?si=jQAFPvy744djWIga

Commented by ajfour last updated on 27/Sep/23

My lecture on You tube Differentiation : Lesson 2

$${My}\:{lecture}\:{on}\:{You}\:{tube}\: \\ $$$${Differentiation}\::\:{Lesson}\:\mathrm{2} \\ $$

Answered by mr W last updated on 27/Sep/23

(h−r)^2 +k^2 =r^2 ⇒k^2 =r^2 −(h−r)^2 =h(2r−h) (h/a^2 )+(k/b^2 )×(−((h−r)/k))=0 ⇒h=((a^2 r)/(a^2 −b^2 ))=(r/(1−μ^2 )) with μ=(b/a) (h^2 /a^2 )+(k^2 /b^2 )=1 (h^2 /a^2 )+((h(2r−h))/b^2 )=1 let λ=(r/a) (λ^2 /((1−μ^2 )^2 ))+((λ^2 (2−(1/(1−μ^2 ))))/(μ^2 (1−μ^2 )))=1 ⇒λ=μ(√(1−μ^2 )) ⇒r=b(√(1−(b^2 /a^2 ))) ✓

$$\left({h}−{r}\right)^{\mathrm{2}} +{k}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\Rightarrow{k}^{\mathrm{2}} ={r}^{\mathrm{2}} −\left({h}−{r}\right)^{\mathrm{2}} ={h}\left(\mathrm{2}{r}−{h}\right) \\ $$$$\frac{{h}}{{a}^{\mathrm{2}} }+\frac{{k}}{{b}^{\mathrm{2}} }×\left(−\frac{{h}−{r}}{{k}}\right)=\mathrm{0} \\ $$$$\Rightarrow{h}=\frac{{a}^{\mathrm{2}} {r}}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }=\frac{{r}}{\mathrm{1}−\mu^{\mathrm{2}} }\:{with}\:\mu=\frac{{b}}{{a}} \\ $$$$\frac{{h}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{k}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{h}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{h}\left(\mathrm{2}{r}−{h}\right)}{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${let}\:\lambda=\frac{{r}}{{a}} \\ $$$$\frac{\lambda^{\mathrm{2}} }{\left(\mathrm{1}−\mu^{\mathrm{2}} \right)^{\mathrm{2}} }+\frac{\lambda^{\mathrm{2}} \left(\mathrm{2}−\frac{\mathrm{1}}{\mathrm{1}−\mu^{\mathrm{2}} }\right)}{\mu^{\mathrm{2}} \left(\mathrm{1}−\mu^{\mathrm{2}} \right)}=\mathrm{1} \\ $$$$\Rightarrow\lambda=\mu\sqrt{\mathrm{1}−\mu^{\mathrm{2}} } \\ $$$$\Rightarrow{r}={b}\sqrt{\mathrm{1}−\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\:\:\checkmark \\ $$

Commented by mr W last updated on 27/Sep/23

Commented by ajfour last updated on 27/Sep/23

Yes sir. Thank you.

Answered by ajfour last updated on 27/Sep/23

(x−r)^2 +y^2 =r^2 (x^2 /a^2 )+(y^2 /b^2 )=1 ⇒ (((x−r)^2 )/b^2 )+1−(x^2 /a^2 )=(r^2 /b^2 ) multiplying by a^2 b^2 (a^2 −b^2 )x^2 −(2ra^2 )x+a^2 b^2 =0 D=0 4r^2 a^4 =4a^2 b^2 (a^2 −b^2 ) r=((b/a))(√(a^2 −b^2 ))

$$\left({x}−{r}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\Rightarrow\:\:\:\frac{\left({x}−{r}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }+\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\frac{{r}^{\mathrm{2}} }{{b}^{\mathrm{2}} } \\ $$$${multiplying}\:{by}\:{a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$$\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){x}^{\mathrm{2}} −\left(\mathrm{2}{ra}^{\mathrm{2}} \right){x}+{a}^{\mathrm{2}} {b}^{\mathrm{2}} =\mathrm{0} \\ $$$${D}=\mathrm{0} \\ $$$$\mathrm{4}{r}^{\mathrm{2}} {a}^{\mathrm{4}} =\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right) \\ $$$${r}=\left(\frac{{b}}{{a}}\right)\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$$$ \\ $$

Commented by mr W last updated on 27/Sep/23

very nice approach!

$${very}\:{nice}\:{approach}! \\ $$

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