Question and Answers Forum

All Questions      Topic List

Coordinate Geometry Questions

Previous in All Question      Next in All Question      

Previous in Coordinate Geometry      Next in Coordinate Geometry      

Question Number 216060 by mr W last updated on 26/Jan/25

Commented by mr W last updated on 26/Jan/25

$$findradiusofinscribedcircler=?$$

Answered by mr W last updated on 26/Jan/25

$$say\mu =\frac{b}{a},\lambda =\frac{r}{a}$$$$sayP(a\cos \theta ,b\sin \theta )$$$$\tan \varphi =-\frac{dx}{dy}=\frac{a\sin \theta }{b\cos \theta }=\frac{\tan \theta }{\mu }=\frac{m}{\mu }$$$$b\sin \theta =r\sin \varphi$$$$\Rightarrow \mu \sin \theta =\frac{\lambda m}{\sqrt{m^2+{\mu }^2}}...\left(i\right)$$$$a\cos \theta =r+r\cos \varphi$$$$\Rightarrow \cos \theta =\lambda (1+\frac{\mu }{\sqrt{m^2+{\mu }^2}})...\left(ii\right)$$$$\left(i\right)/\left(ii\right):$$$$\mu \tan \theta =\frac{m}{\mu +\sqrt{m^2+{\mu }^2}}$$$$\mu =\frac{1}{\mu +\sqrt{m^2+{\mu }^2}}$$$$\sqrt{m^2+{\mu }^2}=\frac{1}{\mu }-\mu$$$$\Rightarrow m^2=\frac{1}{{\mu }^2}-2⩾0\Rightarrow \mu ⩽\frac{1}{\sqrt{2}}\Rightarrow a⩾\sqrt{2}b$$$$from\left(i\right):$$$$\mu \sin \theta =\frac{\lambda m}{\sqrt{m^2+{\mu }^2}}$$$$\frac{\mu m}{\sqrt{1+m^2}}=\frac{\lambda m}{\sqrt{m^2+{\mu }^2}}$$$$\Rightarrow \lambda =\frac{\mu \sqrt{m^2+{\mu }^2}}{\sqrt{1+m^2}}=\frac{\mu (\frac{1}{\mu }-\mu )}{\sqrt{\frac{1}{{\mu }^2}-1}}=\mu \sqrt{1-{\mu }^2}$$$$i.e.r=b\sqrt{1-{\left(\frac{b}{a}\right)}^2}$$

Commented by mr W last updated on 26/Jan/25

Commented by mr W last updated on 26/Jan/25

Commented by ajfour last updated on 27/Jan/25

$$Ifa⩽\sqrt{2}bthensuchr=\frac{a}{2}$$$$Andifa⩾\sqrt{2}br=\frac{b}{a}\sqrt{a^2-b^2}$$$$Isntitthistosummarize,Sir?Thanks.$$

Commented by mr W last updated on 27/Jan/25

$$ifa=\sqrt{2}bthenr=\frac{a}{2}.$$$$ifa<\sqrt{2}bthennoinscribedcircle$$$$possible.$$

Answered by aleks041103 last updated on 27/Jan/25

Commented by aleks041103 last updated on 27/Jan/25

$$Thecircleistangenttotheellipseandtherefore$$$$theradiusis\mathrm{\perp }tothetangenttotheellipse$$$$atthispoint.$$$$Thetangentofanellipseis\mathrm{\perp }totheangle$$$$bisectoroftheanglebetweenthesegments$$$$connectingpointthefocitothepoint.$$$$Thuswegetthetopschematic.$$$$Now:$$$$x+y=2a\Rightarrow y=2a-x$$$$\frac{x}{y}=\frac{c+r}{c-r}$$$$xy-(c+r)(c-r)=r^2\Rightarrow xy=c^2=a^2-b^2$$$$c=a\sqrt{1-{\left(\frac{b}{a}\right)}^2}$$$$\Rightarrow x(2a-x)=a^2-b^2$$$$\Rightarrow {(x-a)}^2-b^2=0$$$$\Rightarrow x=a+b;a-b$$$$butx>y\Rightarrow x=a+b,y=a-b$$$$\Rightarrow \frac{x}{h}=\frac{a+b}{a-b}=\frac{c+r}{c-r}$$$$\Rightarrow r=\frac{bc}{a}$$$$\Rightarrow r=b\sqrt{1-{\left(\frac{b}{a}\right)}^2}$$

Commented by aleks041103 last updated on 27/Jan/25

$$TheradiustothepointP\left(originalschematic\right)$$$$isperpendicular\left(\mathrm{\perp }\right)tothetangentofthe$$$$ellipseatP(sincethetangentofthecircle$$$$coinsideswiththetangenttotheellipse).$$

Commented by mr W last updated on 27/Jan/25

$$thanksforthisnicesolution!$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com