Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 42948 by ajfour last updated on 05/Sep/18

Commented by ajfour last updated on 05/Sep/18

$$FindcoordinatesofpointP,if$$$$thecolouredareaisaminimum.$$

Commented by MJS last updated on 05/Sep/18

$$\mathrm{this}\mathrm{is}\mathrm{not}\mathrm{very}\mathrm{hard},{\mathrm{I}}^{\prime }\mathrm{ll}\mathrm{post}\mathrm{the}\mathrm{answer}\mathrm{later}$$$$\mathrm{if}\mathrm{noone}\mathrm{else}\mathrm{will}\mathrm{have}\mathrm{done}\mathrm{it}...$$$$\mathrm{but}\mathrm{now}\mathrm{find}$$$$\left(1\right)\mathrm{the}\mathrm{greatest}\mathrm{triangle}$$$$\left(2\right)\mathrm{the}\mathrm{greatest}\mathrm{circle}$$$$\left(3\right)\mathrm{the}\mathrm{greatest}\mathrm{trapezoid}$$$$\mathrm{fitting}\mathrm{into}\mathrm{the}\mathrm{found}\mathrm{blue}\mathrm{area}$$

Commented by malwaan last updated on 06/Sep/18

$$\mathrm{please}\mathrm{solve}\mathrm{it}$$

Commented by ajfour last updated on 06/Sep/18

Commented by ajfour last updated on 06/Sep/18

$$\left(2\right)Radiusofgreatestcircle$$$$P(\frac{-1}{2},\frac{1}{4})$$$$slopeofnormalatP:m=1$$$$\Rightarrow y=x+c$$$$\Rightarrow \frac{1}{4}=-\frac{1}{2}+c\Rightarrow c=\frac{3}{4}$$$$letB(x_2,x_2^2)$$$$\Rightarrow formaximumradius$$$$2x_2=1\Rightarrow x_2=\frac{1}{2}$$$$2r=\frac{\mid y_2-x_2-c\mid }{\sqrt{2}}=\frac{\mid \frac{1}{4}-\frac{1}{2}-\frac{3}{4}\mid }{\sqrt{2}}$$$$\Rightarrow {\mathit{r}}_{\mathit{m}\mathit{a}\mathit{x}}=\frac{1}{2\sqrt{2}}.$$

Answered by ajfour last updated on 05/Sep/18

$$leteq.ofnormalbe$$$$y=mx+c$$$$forpointsofintersection$$$$x^2=mx+c$$$$\Rightarrow x_1+x_2=m$$$$and{x}_1{x}_2=-c$$$$further2{mx}_1=-1$$$$\Rightarrow m=-\frac{1}{2x_1},x_2=-\frac{1}{2x_1}-x_1$$$$c=\frac{1}{2}+x_1^2$$$$\Rightarrow \frac{{dx}_2}{{dx}_1}=\frac{1}{2x_1^2}-1$$$$AreaA={\int }_{x_1}^{x_2}(mx+c-x^2)dx$$$$\frac{dA}{{dx}_1}=0\Rightarrow {\int }_{x_1}^{x_2}(x\frac{dm}{{dx}_1}+\frac{dc}{{dx}_1})dx+$$$$\frac{{dx}_2}{{dx}_1}({mx}_2+c-x_2^2)-({mx}_1+c-x_1^2)=0$$$$\Rightarrow {\int }_{x_1}^{x_2}(\frac{x}{2x_1^2}+2x_1)dx$$$$+(\frac{1}{2x_1^2}-1)(\frac{1}{4x_1^2}+\frac{1}{2}+\frac{1}{2}+x_1^2-\frac{1}{4x_1^2}-x_1^2-1)$$$$-(-\frac{1}{2}+\frac{1}{2}+x_1^2-x_1^2)=0$$$$\Rightarrow {\int }_{x_1}^{x_2}(\frac{x}{2x_1^2}+2x_1)dx=0$$$$\Rightarrow \frac{(x_2^2-x_1^2)}{4x_1^2}+2x_1(x_2-x_1)=0$$$$\Rightarrow \frac{x_2+x_1}{4x_1^2}=-2x_1$$$$or-\frac{1}{2x_1}=-8x_1^3$$$$\Rightarrow x_1^2=\frac{1}{4}\Rightarrow \mathit{P}(\pm \frac{1}{2},\frac{1}{4}).$$$$$$

Commented by ajfour last updated on 05/Sep/18

$$Haveisolveditcorrect,Sir?$$

Commented by MJS last updated on 05/Sep/18

$$\mathrm{yes}.\mathrm{with}y={ax}^2\mathrm{we}\mathrm{get}p=\pm \frac{1}{2a}\mathrm{and}\mathrm{blue}$$$$\mathrm{area}=\frac{4}{3a^2}$$

Commented by ajfour last updated on 05/Sep/18

$$ThanksSir.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com