Question Number 19021 by ajfour last updated on 03/Aug/17 Commented by ajfour last updated on 03/Aug/17 $$\mathrm{If}\:\:\:\:\:\:\phi=\mathrm{tan}^{−\mathrm{1}} \left[\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\mathrm{cot}\:\theta\right]+\theta \\ $$$$\Rightarrow\:\mathrm{tan}\:\left(\phi−\theta\right)=\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\mathrm{cot}\:\theta \\ $$$$\:\:\:\frac{\mathrm{tan}\:\phi−\mathrm{tan}\:\theta}{\mathrm{1}+\mathrm{tan}\:\phi\mathrm{tan}\:\theta}=\frac{\mathrm{cot}\:\theta}{\:\sqrt{\mathrm{2}}+\mathrm{1}} \\ $$$$\Rightarrow\:\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)\:\left(\mathrm{tan}\:\phi−\mathrm{tan}\:\theta\right) \\…
Question Number 84531 by jagoll last updated on 14/Mar/20 $$\mathrm{find}\:\mathrm{for}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{image}\:\mathrm{ellipse} \\ $$$$\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{9}}\:+\:\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{8}}\:=\:\mathrm{1}\:\mathrm{if}\:\mathrm{reflected}\:\mathrm{with}\:\mathrm{line} \\ $$$$\mathrm{x}\:+\:\mathrm{y}\:=\:−\mathrm{4} \\ $$ Commented by jagoll last updated on 14/Mar/20…
Question Number 150056 by ajfour last updated on 09/Aug/21 Commented by ajfour last updated on 09/Aug/21 $${If}\:{the}\:{blue}\:{region}\:{is}\:{a}\:{square} \\ $$$${of}\:{maximum}\:{area},\:{find}\:{a} \\ $$$${and}\:{side}\:{s}\:{of}\:{the}\:{square}.\:\:\: \\ $$ Answered by…
Question Number 84512 by 698148290 last updated on 13/Mar/20 Answered by jagoll last updated on 14/Mar/20 $$\mathrm{tangent}\:\mathrm{equation}\:\mathrm{at}\:\mathrm{point} \\ $$$$\mathrm{P}\left(\mathrm{2a}+\mathrm{2t}\:,\:\frac{\mathrm{at}^{\mathrm{2}} }{\mathrm{2}}\right)\:\mathrm{to}\:\mathrm{the}\:\mathrm{parabola} \\ $$$$\left(\mathrm{x}−\mathrm{2a}\right)^{\mathrm{2}} \:=\:\mathrm{2ay}\: \\ $$$$\Rightarrow\:\mathrm{2t}\left(\mathrm{x}−\mathrm{2a}\right)\:=\:\mathrm{ay}\:+\:\frac{\left(\mathrm{at}\right)^{\mathrm{2}}…
Question Number 84441 by Power last updated on 13/Mar/20 Commented by Power last updated on 13/Mar/20 $$\mathrm{sir}\:\mathrm{please}\:\mathrm{solution} \\ $$ Answered by mr W last updated…
Question Number 18904 by Satyamtt last updated on 01/Aug/17 $${Solve}\:{the}\:{triangle}\:{in}\:{which}\:{a}=\left(\sqrt{\mathrm{3}}+\mathrm{1}\right),\: \\ $$$${b}=\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\:{and}\:\angle{C}=\mathrm{60}°. \\ $$$$ \\ $$ Answered by behi.8.3.4.1.7@gmail.com last updated on 01/Aug/17 $${c}^{\mathrm{2}} ={a}^{\mathrm{2}}…
Question Number 18823 by rish@bh last updated on 30/Jul/17 $$\mathrm{If}\:\mathrm{P}\equiv\left(\mathrm{2},\mathrm{1}\right)\:\mathrm{and}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{lie}\:\mathrm{on}\:\mathrm{x}−\mathrm{axis} \\ $$$$\mathrm{and}\:\mathrm{y}=\mathrm{x}\:\mathrm{respectively}\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\mathrm{PA}+\mathrm{PB}+\mathrm{AB}\:\mathrm{is}\:\mathrm{minimum}\:\mathrm{find} \\ $$$$\mathrm{A}\:\mathrm{and}\:\mathrm{B}. \\ $$ Commented by ajfour last updated on 30/Jul/17…
Question Number 149796 by mr W last updated on 07/Aug/21 Commented by mr W last updated on 07/Aug/21 $${solution}\:{to}\:{Q}\mathrm{148806} \\ $$ Commented by mr W…
Question Number 18642 by tawa tawa last updated on 26/Jul/17 $$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{region}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{coordinate}\:\mathrm{axis}\:\mathrm{and}\:\mathrm{the}\:\mathrm{line} \\ $$$$\mathrm{tangent}\:\mathrm{to}\:\mathrm{the}\:\mathrm{graph},\:\:\mathrm{y}\:=\:\frac{\mathrm{1}}{\mathrm{8}}\mathrm{x}^{\mathrm{2}} \:+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}\:+\:\mathrm{1},\:\mathrm{at}\:\mathrm{the}\:\mathrm{point}\:\:\left(\mathrm{0},\:\mathrm{1}\right) \\ $$ Answered by Tinkutara last updated on 26/Jul/17 $$\frac{{dy}}{{dx}}\:=\:\frac{{x}}{\mathrm{4}}\:+\:\frac{\mathrm{1}}{\mathrm{2}} \\…
Question Number 84157 by redmiiuser last updated on 10/Mar/20 $${if}\:{a}\:{circle}\:{having}\:{an}\: \\ $$$${equation}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{6}{x}−\mathrm{8}{y}=\mathrm{0} \\ $$$${is}\:{intersected}\:{at}\:{A}\:{and}\:{B} \\ $$$${by}\:{x}+{y}=\mathrm{1}.{find}\:{the}\: \\ $$$${equation}\:{of}\:{the}\:{circle} \\ $$$${on}\:{AB}\:{as}\:{diameter} \\ $$ Answered…