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Question Number 48359 by ajfour last updated on 22/Nov/18

Answered by mr W last updated on 23/Nov/18

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$$$\frac{x}{a^2}+\frac{{yy}^{\prime }}{b^2}=0\Rightarrow y^{\prime }=-\frac{b^{2}x}{a^{2}y}$$$${(x-r)}^2+{(y-r)}^2=r^2$$$$(x-r)+(y-r)y^{\prime }=0\Rightarrow y^{\prime }=-\frac{x-r}{y-r}$$$$T(h,k)$$$$\frac{h^2}{a^2}+\frac{k^2}{b^2}=1$$$${(h-r)}^2+{(k-r)}^2=r^2$$$$\frac{b^{2}h}{a^{2}k}=\frac{h}{b+k}=\frac{r}{b+r}$$$$\Rightarrow b^2(b+k)=a^{2}k\Rightarrow k=\frac{b^3}{a^2-b^2}$$$$\frac{h^2}{a^2}=1-\frac{b^6}{b^2{(a^2-b^2)}^2}=1-{\left(\frac{b^2}{a^2-b^2}\right)}^2=\frac{a^2(a^2-2b^2)}{(a^2-b^2)}$$$$\Rightarrow h=\frac{a^2\sqrt{a^2-2b^2}}{a^2-b^2}$$$$(b+k-h)r=bh$$$$\Rightarrow r=\frac{bh}{b+k-h}=\frac{{ba}^2\sqrt{a^2-2b^2}}{(a^2-b^2)(b+\frac{b^3}{a^2-b^2}-\frac{a^2\sqrt{a^2-2b^2}}{a^2-b^2})}=\frac{b\sqrt{a^2-2b^2}}{b-\sqrt{a^2-2b^2}}$$$$with\lambda =\frac{a}{b}$$$$\Rightarrow k=\frac{b}{{\lambda }^2-1}$$$$\Rightarrow h=\frac{b{\lambda }^2\sqrt{{\lambda }^2-2}}{{\lambda }^2-1}$$$$\Rightarrow r=\frac{b\sqrt{{\lambda }^2-2}}{1-\sqrt{{\lambda }^2-2}}$$$$\Rightarrow \frac{h}{r}=\frac{{\lambda }^2(1-\sqrt{{\lambda }^2-2})}{{\lambda }^2-1}$$$$\Rightarrow \frac{k}{r}=\frac{1-\sqrt{{\lambda }^2-2}}{({\lambda }^2-1)\sqrt{{\lambda }^2-2}}$$$${(\frac{h}{r}-1)}^2+{(\frac{k}{r}-1)}^2=1$$$${[\frac{{\lambda }^2(1-\sqrt{{\lambda }^2-2})}{{\lambda }^2-1}-1]}^2+{[\frac{1-\sqrt{{\lambda }^2-2}}{({\lambda }^2-1)\sqrt{{\lambda }^2-2}}-1]}^2=1$$$${\left[\frac{1-{\lambda }^2\sqrt{{\lambda }^2-2}}{{\lambda }^2-1}\right]}^2+{\left[\frac{1-{\lambda }^2\sqrt{{\lambda }^2-2}}{({\lambda }^2-1)\sqrt{{\lambda }^2-2}}\right]}^2=1$$$$\frac{1-{\lambda }^2\sqrt{{\lambda }^2-2}}{\sqrt{({\lambda }^2-1)({\lambda }^2-2)}}=1$$$$\Rightarrow ({\lambda }^2+\sqrt{{\lambda }^2-1})\sqrt{{\lambda }^2-2}=1$$$$\Rightarrow \lambda =\frac{a}{b}\approx 1.4494$$

Commented by ajfour last updated on 23/Nov/18

$$Great!$$

Commented by mr W last updated on 23/Nov/18

$$toajfoursir:$$$$-y^{\prime }=\frac{b^{2}h}{a^{2}k}=\frac{h-r}{k-r}=\frac{h}{b+k}=\frac{r}{b+r}$$$$thecondition=\frac{h-r}{k-r}isnotnecessary,$$$$because{it}^{\prime }sincludedin=\frac{h}{b+k}=\frac{r}{b+r}.$$

Commented by ajfour last updated on 23/Nov/18

$$how\frac{h-R}{k-R}=\frac{h}{b+k}Sir?$$$$Line8.$$

Commented by MJS last updated on 23/Nov/18

$${\mathrm{you}}^{\prime }\mathrm{re}\mathrm{losing}\mathrm{the}{2}^{\mathrm{nd}}\mathrm{solution}\mathrm{by}\mathrm{taking}\mathrm{the}$$$$\mathrm{root}\mathrm{in}\mathrm{the}{3}^{\mathrm{rd}}\mathrm{line}\mathrm{from}\mathrm{the}\mathrm{bottom}.$$$$\lambda =\pm 1.44941\vee \lambda =\pm 1.60446$$$$\Rightarrow 2\mathrm{circles}\mathrm{are}\mathrm{possible}$$

Commented by mr W last updated on 23/Nov/18

$$toMJSsir:$$$$youareright.thefinaleqn.aftertaking$$$$rootshouldbe$$$$\frac{1-{\lambda }^2\sqrt{{\lambda }^2-2}}{\sqrt{({\lambda }^2-1)({\lambda }^2-2)}}=\pm 1$$$$or$$$$({\lambda }^2\pm \sqrt{{\lambda }^2-1})\sqrt{{\lambda }^2-2}=1$$$$but‘‘{-}^{″}isforthecircleoutsideofellipse.$$

Commented by mr W last updated on 23/Nov/18

Commented by mr W last updated on 23/Nov/18

Commented by ajfour last updated on 23/Nov/18

$$Thankyouboth.Ishouldhave$$$$drawntwosuchcirclesinthe$$$$questiondiagramitself..$$

Answered by MJS last updated on 22/Nov/18

$$T=\left(\begin{array}{c}\frac{a^2\sqrt{a^2-2b^2}}{a^2-b^2}\\ \frac{b^3}{a^2-b^2}\end{array}\right)B=\left(\begin{array}{c}0\\ -b\end{array}\right)$$$$\mathrm{from}C\in y=x\wedge C\in BT\Rightarrow$$$$\Rightarrow C=\left(\begin{array}{c}r\\ r\end{array}\right)r=\frac{b\sqrt{a^2-2b^2}}{b-\sqrt{a^2-2b^2}}$$$$$$$$\mathrm{I}\mathrm{put}a=1$$$$\mid CT{\mid }^2=r^2$$$$\mathrm{this}\mathrm{leads}\mathrm{to}\mathrm{a}{6}^{\mathrm{th}}\mathrm{degree}\mathrm{polynome}\mathrm{which}\mathrm{had}$$$$2\mathrm{real}\mathrm{solutions}:$$$$a=1;b_1=.689937;c_1=1.95025$$$$\left[\mathrm{circle}\mathrm{touches}\mathrm{ellipse}\mathrm{from}\mathrm{outside}\right]$$$$a=1;b_2=.623263;c_2=.320906$$$$\left[\mathrm{circle}\mathrm{touches}\mathrm{ellipse}\mathrm{from}\mathrm{inside}\right]$$

Commented by ajfour last updated on 23/Nov/18

$$Sir,firstofallhowtoarriveat$$$$coordinatesofT;yourmethod?$$

Commented by MJS last updated on 23/Nov/18

$$y=\frac{b}{a}\sqrt{a^2-x^2}$$$$\mathrm{normal}\mathrm{in}x=t:y=kx+d$$$$y^{\prime }=-\frac{bt}{a\sqrt{a^2-t^2}}\Rightarrow k=\frac{a\sqrt{a^2-t^2}}{bt}$$$$d=y-kt=-\frac{a^2-b^2}{ab}\sqrt{a^2-t^2}$$$$\mathrm{but}d=-b\Rightarrow t=\pm \frac{a^2\sqrt{a^2-2b^3}}{a^2-b^2}\wedge a>b\wedge a^2⩾2b^2$$$$u=\frac{b}{a}\sqrt{a^2-t^2}=\frac{b^3}{a^2-b^2}$$$$\mathrm{line}BT:y=\frac{b+u}{t}x-b$$$$(y=\frac{b+u}{t}x-b)\cap (y=x)\Rightarrow x=\frac{bt}{b-t+u}=\frac{b\sqrt{a^2-2b^2}}{b-\sqrt{a^2-2b^2}}=r$$$$\mid CT{\mid }^2=r^2$$$$\mathrm{this}\mathrm{leads}\mathrm{to}$$$$2a^2{b}^3\sqrt{a^2-2b^2}={(a-b)}^3{(a+b)}^3$$$$\mathrm{squaring}\mathrm{leads}\mathrm{to}$$$$a^{12}-6a^{10}{b}^2+15a^8{b}^4-24a^6{b}^6+23a^4{b}^8-6a^2{b}^{10}+b^{12}=0$$$$a=\sqrt{p};b=\sqrt{q}$$$$p^6-6p^{5}q+15p^4{q}^2-24p^3{q}^3+23p^2{q}^4-6{pq}^5+q^6=0$$$$p=1$$$$q^6-6q^5+23q^4-24q^3+15q^2-6q+1=0$$$$\mathrm{approximating}\Rightarrow q=.388457\vee q=.476013$$$$\Rightarrow b=.623263\vee b=.689937$$

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