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

Commented by ajfour last updated on 27/Sep/23

$$Determinerintermsofa,b.$$

Commented by ajfour last updated on 27/Sep/23

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

Commented by ajfour last updated on 27/Sep/23

$$MylectureonYoutube$$$$Differentiation:Lesson2$$

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

$${(h-r)}^2+k^2=r^2$$$$\Rightarrow k^2=r^2-{(h-r)}^2=h(2r-h)$$$$\frac{h}{a^2}+\frac{k}{b^2}\times (-\frac{h-r}{k})=0$$$$\Rightarrow h=\frac{a^{2}r}{a^2-b^2}=\frac{r}{1-{\mu }^2}with\mu =\frac{b}{a}$$$$\frac{h^2}{a^2}+\frac{k^2}{b^2}=1$$$$\frac{h^2}{a^2}+\frac{h(2r-h)}{b^2}=1$$$$let\lambda =\frac{r}{a}$$$$\frac{{\lambda }^2}{{(1-{\mu }^2)}^2}+\frac{{\lambda }^2(2-\frac{1}{1-{\mu }^2})}{{\mu }^2(1-{\mu }^2)}=1$$$$\Rightarrow \lambda =\mu \sqrt{1-{\mu }^2}$$$$\Rightarrow r=b\sqrt{1-\frac{b^2}{a^2}}✓$$

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

$${(x-r)}^2+y^2=r^2$$$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$$$\Rightarrow \frac{{(x-r)}^2}{b^2}+1-\frac{x^2}{a^2}=\frac{r^2}{b^2}$$$$multiplyingby{a}^2{b}^2$$$$(a^2-b^2)x^2-\left(2{ra}^2\right)x+a^2{b}^2=0$$$$D=0$$$$4r^2{a}^4=4a^2{b}^2(a^2-b^2)$$$$r=\left(\frac{b}{a}\right)\sqrt{a^2-b^2}$$$$$$

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

$$veryniceapproach!$$

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