Question Number 197730 by ajfour last updated on 27/Sep/23
Commented by ajfour last updated on 27/Sep/23
$${Determine}\:{r}\:{in}\:{terms}\:{of}\:{a},{b}. \\ $$
Commented by ajfour last updated on 27/Sep/23
https://youtu.be/0sSIzDZtOBQ?si=jQAFPvy744djWIga
Commented by ajfour last updated on 27/Sep/23
$${My}\:{lecture}\:{on}\:{You}\:{tube}\: \\ $$$${Differentiation}\::\:{Lesson}\:\mathrm{2} \\ $$
Answered by mr W last updated on 27/Sep/23
$$\left({h}−{r}\right)^{\mathrm{2}} +{k}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\Rightarrow{k}^{\mathrm{2}} ={r}^{\mathrm{2}} −\left({h}−{r}\right)^{\mathrm{2}} ={h}\left(\mathrm{2}{r}−{h}\right) \\ $$$$\frac{{h}}{{a}^{\mathrm{2}} }+\frac{{k}}{{b}^{\mathrm{2}} }×\left(−\frac{{h}−{r}}{{k}}\right)=\mathrm{0} \\ $$$$\Rightarrow{h}=\frac{{a}^{\mathrm{2}} {r}}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }=\frac{{r}}{\mathrm{1}−\mu^{\mathrm{2}} }\:{with}\:\mu=\frac{{b}}{{a}} \\ $$$$\frac{{h}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{k}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{h}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{h}\left(\mathrm{2}{r}−{h}\right)}{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${let}\:\lambda=\frac{{r}}{{a}} \\ $$$$\frac{\lambda^{\mathrm{2}} }{\left(\mathrm{1}−\mu^{\mathrm{2}} \right)^{\mathrm{2}} }+\frac{\lambda^{\mathrm{2}} \left(\mathrm{2}−\frac{\mathrm{1}}{\mathrm{1}−\mu^{\mathrm{2}} }\right)}{\mu^{\mathrm{2}} \left(\mathrm{1}−\mu^{\mathrm{2}} \right)}=\mathrm{1} \\ $$$$\Rightarrow\lambda=\mu\sqrt{\mathrm{1}−\mu^{\mathrm{2}} } \\ $$$$\Rightarrow{r}={b}\sqrt{\mathrm{1}−\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\:\:\checkmark \\ $$
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
$$\left({x}−{r}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\Rightarrow\:\:\:\frac{\left({x}−{r}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }+\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\frac{{r}^{\mathrm{2}} }{{b}^{\mathrm{2}} } \\ $$$${multiplying}\:{by}\:{a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$$\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){x}^{\mathrm{2}} −\left(\mathrm{2}{ra}^{\mathrm{2}} \right){x}+{a}^{\mathrm{2}} {b}^{\mathrm{2}} =\mathrm{0} \\ $$$${D}=\mathrm{0} \\ $$$$\mathrm{4}{r}^{\mathrm{2}} {a}^{\mathrm{4}} =\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right) \\ $$$${r}=\left(\frac{{b}}{{a}}\right)\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$$$ \\ $$
Commented by mr W last updated on 27/Sep/23
$${very}\:{nice}\:{approach}! \\ $$