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 64436 by ajfour last updated on 17/Jul/19

Commented by ajfour last updated on 17/Jul/19

$$Findradiusofarclengths,if$$$$areabetweenparabolaandarc$$$$isamaximum.$$

Answered by Tanmay chaudhury last updated on 18/Jul/19

$$y=x^{2}curvepassesthrough(\alpha ,{\alpha }^2)and(-\alpha ,{\alpha }^2)$$$$arcisaportionofcircle...thatcirclepass$$$$through(\alpha ,{\alpha }^2),(-\alpha ,{\alpha }^2)and(0,\beta )$$$$x^2+y^2+2gx+2fy+c=0$$$${\alpha }^2+{\alpha }^4+2g\alpha +2f{\alpha }^2+c=0$$$${\alpha }^2+{\alpha }^4-2g\alpha +2f{\alpha }^2+c=0$$$${\beta }^2+2f\beta +c=0$$$$4g\alpha =0g=0$$$${\alpha }^2+{\alpha }^4+2f{\alpha }^2+c=0$$$${\beta }^2+2f\beta +c=0$$$$2f({\alpha }^2-\beta )+{\alpha }^2+{\alpha }^4-{\beta }^2=0$$$$f=\frac{{\beta }^2-{\alpha }^2-{\alpha }^4}{2({\alpha }^2-\beta )}$$$${\beta }^2+\left(\frac{{\beta }^2-{\alpha }^2-{\alpha }^4}{{\alpha }^2-\beta }\right)\beta +c=0$$$$\frac{{\alpha }^2{\beta }^2-{\beta }^3+{\beta }^3-\beta ({\alpha }^2+{\alpha }^4)}{{\alpha }^2-\beta }+c=0$$$$\frac{{\alpha }^2\beta (\beta -1-{\alpha }^2)}{{\alpha }^2-\beta }+c=0$$$$c=\frac{{\alpha }^2\beta (1+{\alpha }^2-\beta )}{{\alpha }^2-\beta }$$$$eqnofcircle$$$$x^2+y^2+2\left(\frac{{\beta }^2-{\alpha }^2-{\alpha }^4}{2({\alpha }^2-\beta )}\right)y+\frac{{\alpha }^2\beta (1+{\alpha }^2-\beta )}{{\alpha }^2-\beta }=0$$$$x^2+y^2+2Py+Q=0$$$$Radius=\sqrt{P^2-Q}$$$$y^2+2Py+P^2+Q+x^2=P^2$$$${(y+P)}^2=P^2-Q-x^2$$$$y=-P+\sqrt{P^2-Q-x^2}$$$$requiredareawhichtomaximise$$$$2[{\int }_0^{{\alpha }^2}[(-P+\sqrt{P^2-Q-x^2})-x^2]dx$$$$2\mid Px-\frac{x^3}{3}+\frac{x}{2}\sqrt{P^2-Q-x^2}+\frac{P^2-Q}{2}{sin}^{-1}\left(\frac{x}{\sqrt{P^2-Q}}\right){\mid }_0^{{\alpha }^2}$$$$2[P{\alpha }^2-\frac{{\alpha }^6}{3}+\frac{{\alpha }^2}{2}\sqrt{P^2-Q-{\alpha }^4}+\frac{P^2-Q}{2}{sin}^{-1}\left(\frac{{\alpha }^2}{\sqrt{P^2-Q}}\right)$$$$plscheckuptothis...nxtstepputPandQ$$$$maximizearea$$$$thenfinallyR$$

Answered by ajfour last updated on 18/Jul/19

$$letcircleeq.be{\mathit{x}}^2+{(\mathit{y}-\mathit{c})}^2={\mathit{r}}^2$$$$Alsolet\mathit{r}\mathit{\theta }=\frac{\mathit{s}}{2}$$$$Intersectionofparabolaand$$$$circleontherightbe\mathit{P}(\mathit{h},{\mathit{h}}^2).$$$$h^2=c+r\cos \theta ,h=r\sin \theta$$$$\frac{\mathit{A}}{2}={\int }_0^{h^2}\sqrt{y}dy-\frac{r^2\sin \theta \cos \theta }{2}+\frac{r^2\theta }{2}$$$$A=\frac{4h^3}{3}-r^2\sin \theta \cos \theta +r^2\theta$$$$A={\left(\frac{s}{2\theta }\right)}^2[\frac{2s}{3\theta }\sin {}^3\theta -\sin \theta \cos \theta +\theta ]$$$$\frac{dA}{d\theta }=-\frac{s^2}{2{\theta }^3}[\frac{2s}{3\theta }\sin {}^3\theta -\sin \theta \cos \theta +\theta ]$$$$+\frac{s^2}{4{\theta }^2}[-\frac{2s}{3{\theta }^3}\sin {}^3\theta +\frac{2s}{\theta }\sin {}^2\theta -\cos {}^2\theta$$$$+\sin {}^2\theta +1]$$$$\Rightarrow \frac{4s}{3\theta }\sin {}^3\theta -2\sin \theta \cos \theta +2\theta$$$$=-\frac{2s}{3{\theta }^2}+2s\sin {}^2\theta -\theta \cos {}^2\theta +\theta \sin {}^2\theta +\theta$$$$.....$$

Commented by ajfour last updated on 18/Jul/19

$$cananycalculatorsolvethiseq.$$$$andprovide\theta intermsof\mathit{s}?$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com