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 51181 by ajfour last updated on 24/Dec/18

Commented by ajfour last updated on 24/Dec/18

$$Findparametersaandbofmaximum$$$$areaellipsewithinsectorofradius$$$$\mathit{r}andcentralangle\mathit{\alpha }.$$

Answered by mr W last updated on 24/Dec/18

Commented by mr W last updated on 25/Dec/18

$$eqn.ofellipse:$$$$\frac{x^2}{a^2}+\frac{{(y-c)}^2}{b^2}=1$$$$eqn.ofOQ:$$$$y=x\tan (\frac{\pi }{2}-\frac{\alpha }{2})=\frac{x}{\tan \frac{\alpha }{2}}=mx$$$$\Rightarrow a^2{m}^2+b^2=c^2$$$$eqn.ofcircle:$$$$x^2+y^2=r^2$$$$\frac{r^2-y^2}{a^2}+\frac{{(y-c)}^2}{b^2}=1$$$$$$$$case1:\alpha ⩽90°$$$$\frac{{(r-c)}^2}{b^2}=1$$$$\Rightarrow b^2={(r-c)}^2$$$$\Rightarrow a^2{m}^2=c^2-{(r-c)}^2=r(2c-r)$$$$P=a^2{b}^2{m}^2=r(2c-r){(r-c)}^2$$$$\frac{dP}{dc}=0$$$$2{(r-c)}^2-(2c-r)2(r-c)=0$$$$\Rightarrow c=\frac{2r}{3}$$$$\Rightarrow b^2=\frac{r^2}{9}$$$$\Rightarrow a^2=\frac{c^2-b^2}{m^2}=\frac{r^2}{3m^2}$$$$$$$$case2:90°⩽\alpha ⩽180°$$$$b^2{r}^2-b^2{y}^2+a^2(y^2-2cy+c^2)=a^2{b}^2$$$$(a^2-b^2)y^2-2a^{2}cy+a^2{c}^2+b^2(r^2-a^2)=0$$$$D=4a^4{c}^2-4(a^2-b^2)(a^2{c}^2+b^2{r}^2-a^2{b}^2)=0$$$$\Rightarrow (a^2-b^2)(r^2-a^2)=a^2{c}^2$$$$\Rightarrow (a^2-b^2)(r^2-a^2)=a^2(a^2{m}^2+b^2)$$$$letP=a^2{b}^2$$$$\Rightarrow (a^2-\frac{P}{a^2})(r^2-a^2)=a^2(a^2{m}^2+\frac{P}{a^2})$$$$\Rightarrow (a^4-P)(r^2-a^2)=a^2(a^4{m}^2+P)$$$$\frac{dP}{da}=0$$$$(a^4-P)(-2a)+\left(4a^3\right)(r^2-a^2)=2a(a^4{m}^2+P)+a^2\left(4a^3{m}^2\right)$$$$3a^2(m^2+1)=2r^2$$$$\Rightarrow a^2=\frac{2r^2}{3(m^2+1)}\Rightarrow a=r\sqrt{\frac{2}{3}}\sin \frac{\alpha }{2}$$$$\Rightarrow (a^2-b^2)(r^2-a^2)=a^2(a^2{m}^2+b^2)$$$$\Rightarrow b^2{r}^2=a^2[r^2-a^2(m^2+1)]$$$$\Rightarrow b^2=\frac{a^2}{3}=\frac{2r^2}{9(m^2+1)}\Rightarrow b=r\frac{\sqrt{2}}{3}\sin \frac{\alpha }{2}$$$$\Rightarrow c^2=a^2{m}^2+b^2=\frac{2(3m^2+1)r^2}{9(m^2+1)}\Rightarrow c=\frac{r\sqrt{2(2+\cos \alpha )}}{3}$$$$condition:c+b⩽r$$$$\frac{r\sqrt{2(2+\cos \alpha )}}{3}+r\frac{\sqrt{2}}{3}\sin \frac{\alpha }{2}⩽r$$$$\Rightarrow \sin \frac{\alpha }{2}⩾\frac{1}{\sqrt{2}}\Rightarrow \frac{\alpha }{2}⩾\frac{\pi }{4}\Rightarrow \alpha ⩾\frac{\pi }{2}$$

Commented by mr W last updated on 24/Dec/18

Commented by mr W last updated on 24/Dec/18

Commented by mr W last updated on 24/Dec/18

Commented by ajfour last updated on 25/Dec/18

$$Detailedandextremely\mathcal{G}ood$$$$solutionSir!Thankyouplentiful.$$

Answered by MJS last updated on 24/Dec/18

$$\mathrm{using}\mathrm{my}\mathrm{beloved}\mathrm{coordinate}\mathrm{method}\mathrm{I}\mathrm{get}$$$$a=\frac{\sqrt{6}}{3}r\sin \frac{\alpha }{2}$$$$b=\frac{\sqrt{2}}{3}r\sin \frac{\alpha }{2}$$

Commented by MJS last updated on 24/Dec/18

$${\mathrm{you}}^{\prime }\mathrm{re}\mathrm{right}.\mathrm{thank}\mathrm{you}\mathrm{for}\mathrm{checking}!$$

Commented by mr W last updated on 24/Dec/18

$$answeriscorrectsir.butthisisonly$$$$forobtuseangle\alpha .$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com