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Question Number 41117 by ajfour last updated on 02/Aug/18

Answered by MJS last updated on 02/Aug/18

r=1 P= ((0),(0) ) A= (((cos α)),((1+sin α)) ) F= (((cos α)),(0) ) ∣AP∣=(√(2(1+sin α))) ∣AF∣=1+sin α ∣FP∣=cos α area=(1/4)(sin 2α +2cos α) (d/dα)[area]=(1/2)(cos 2α −sin α)=0 ⇒ α∈{(π/6), ((5π)/6), ((3π)/2)} area({(π/6), ((5π)/6), ((3π)/2)})={((3(√3))/8), −((3(√3))/8), 0} ⇒ ⇒ α=(π/6), area(AFP)=((3(√3))/8)r^2

$${r}=\mathrm{1} \\ $$$${P}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\:{A}=\begin{pmatrix}{\mathrm{cos}\:\alpha}\\{\mathrm{1}+\mathrm{sin}\:\alpha}\end{pmatrix}\:\:{F}=\begin{pmatrix}{\mathrm{cos}\:\alpha}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\mid{AP}\mid=\sqrt{\mathrm{2}\left(\mathrm{1}+\mathrm{sin}\:\alpha\right)} \\ $$$$\mid{AF}\mid=\mathrm{1}+\mathrm{sin}\:\alpha \\ $$$$\mid{FP}\mid=\mathrm{cos}\:\alpha \\ $$$$\mathrm{area}=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{sin}\:\mathrm{2}\alpha\:+\mathrm{2cos}\:\alpha\right) \\ $$$$\frac{{d}}{{d}\alpha}\left[\mathrm{area}\right]=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\:\mathrm{2}\alpha\:−\mathrm{sin}\:\alpha\right)=\mathrm{0}\:\Rightarrow\:\alpha\in\left\{\frac{\pi}{\mathrm{6}},\:\frac{\mathrm{5}\pi}{\mathrm{6}},\:\frac{\mathrm{3}\pi}{\mathrm{2}}\right\} \\ $$$$\mathrm{area}\left(\left\{\frac{\pi}{\mathrm{6}},\:\frac{\mathrm{5}\pi}{\mathrm{6}},\:\frac{\mathrm{3}\pi}{\mathrm{2}}\right\}\right)=\left\{\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}},\:−\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}},\:\mathrm{0}\right\}\:\Rightarrow \\ $$$$\Rightarrow\:\alpha=\frac{\pi}{\mathrm{6}},\:\mathrm{area}\left({AFP}\right)=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}}{r}^{\mathrm{2}} \\ $$

Commented by ajfour last updated on 02/Aug/18

right answer sir; thanks which place are you from MjS Sir?

$${right}\:{answer}\:{sir};\:{thanks} \\ $$$${which}\:{place}\:{are}\:{you}\:{from}\:{MjS}\:{Sir}? \\ $$

Commented by MJS last updated on 02/Aug/18

I′m from Europe/Austria/Vienna most users here seem to be from India

$$\mathrm{I}'\mathrm{m}\:\mathrm{from}\:\mathrm{Europe}/\mathrm{Austria}/\mathrm{Vienna} \\ $$$$\mathrm{most}\:\mathrm{users}\:\mathrm{here}\:\mathrm{seem}\:\mathrm{to}\:\mathrm{be}\:\mathrm{from}\:\mathrm{India} \\ $$

Commented by ajfour last updated on 02/Aug/18

Nice to know Sir, (Arnold is from Austria too, i believe-movie actor ?)

$${Nice}\:{to}\:{know}\:{Sir},\:\left({Arnold}\:{is}\:{from}\right. \\ $$$$\left.{Austria}\:{too},\:{i}\:{believe}-{movie}\:{actor}\:?\right) \\ $$

Commented by MJS last updated on 02/Aug/18

yes. former body−builder and governer of California... I′m glad to be un−famous musician and hobby−mathematician. it′s a more peaceful life ;−)

$$\mathrm{yes}.\:\mathrm{former}\:\mathrm{body}−\mathrm{builder}\:\mathrm{and}\:\mathrm{governer}\:\mathrm{of} \\ $$$$\mathrm{California}... \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{glad}\:\mathrm{to}\:\mathrm{be}\:\mathrm{un}−\mathrm{famous}\:\mathrm{musician}\:\mathrm{and} \\ $$$$\mathrm{hobby}−\mathrm{mathematician}.\:\mathrm{it}'\mathrm{s}\:\mathrm{a}\:\mathrm{more}\:\mathrm{peaceful} \\ $$$$\left.\mathrm{life}\:;−\right) \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 02/Aug/18

FP=a FA=b area(s)=(1/2)ab o be the centre OP=OA=r FA tangent at P so OPF=90^o let ∠OPA=θ=∠OAP cos(180^o −2θ)=((r^2 +r^2 −PA^2 )/(2r.r)) −2r^2 cos2θ=2r^2 −PA^2 PA^2 =2r^2 +2r^2 cos2θ PA^2 =2r^2 (1+cos2θ) PA=2rcosθ ∠OPF=∠OPA+∠APF=90^o ∠APF=90^o −∠OPA=90^o −θ ∠PAF=θ so sinθ=((PF)/(PA))=(a/(2rcosθ)) a=2rsinθcosθ=rsin2θ cosθ=((AF)/(PA))=(b/(2rcosθ)) so b=2rcos^2 θ^ =r(1+cos2θ) area of triangle=(1/2)ab s =(1/2)rsin2θ.r(1+cos2θ) s=(r^2 /2)(sin2θ)(1+cos2θ) (ds/dθ)=(r^2 /2){sin2θ.−2sin2θ+(1+cos2θ)(cos2θ).2} (ds/dθ)=0 −2sin^2 2θ+2cos2θ+2cos^2 2θ=0 cos4θ+cos2θ=0 2cos3θ.cosθ=0 if cosθ=0 so θ=90^o whichis not feasible so cos3θ=0=cos90^o θ=30^o area max=(r^2 /2)(sin60^o )(1+cos60^o ) =(r^2 /2)(((√3)/2))(1+(1/2))=((3(√3))/8)r^2

$${FP}={a}\:\:{FA}={b}\:\:{area}\left({s}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ab} \\ $$$${o}\:{be}\:{the}\:{centre}\:{OP}={OA}={r}\:\:{FA}\:{tangent}\:{at}\:{P} \\ $$$${so}\:{OPF}=\mathrm{90}^{{o}} \:\:{let}\:\angle{OPA}=\theta=\angle{OAP} \\ $$$${cos}\left(\mathrm{180}^{{o}} −\mathrm{2}\theta\right)=\frac{{r}^{\mathrm{2}} +{r}^{\mathrm{2}} −{PA}^{\mathrm{2}} }{\mathrm{2}{r}.{r}} \\ $$$$−\mathrm{2}{r}^{\mathrm{2}} {cos}\mathrm{2}\theta=\mathrm{2}{r}^{\mathrm{2}} −{PA}^{\mathrm{2}} \\ $$$${PA}^{\mathrm{2}} =\mathrm{2}{r}^{\mathrm{2}} +\mathrm{2}{r}^{\mathrm{2}} {cos}\mathrm{2}\theta \\ $$$${PA}^{\mathrm{2}} =\mathrm{2}{r}^{\mathrm{2}} \left(\mathrm{1}+{cos}\mathrm{2}\theta\right) \\ $$$${PA}=\mathrm{2}{rcos}\theta \\ $$$$\angle{OPF}=\angle{OPA}+\angle{APF}=\mathrm{90}^{{o}} \\ $$$$\angle{APF}=\mathrm{90}^{{o}} −\angle{OPA}=\mathrm{90}^{{o}} −\theta \\ $$$$\angle{PAF}=\theta\:\:{so}\:{sin}\theta=\frac{{PF}}{{PA}}=\frac{{a}}{\mathrm{2}{rcos}\theta} \\ $$$${a}=\mathrm{2}{rsin}\theta{cos}\theta={rsin}\mathrm{2}\theta \\ $$$${cos}\theta=\frac{{AF}}{{PA}}=\frac{{b}}{\mathrm{2}{rcos}\theta}\:\:{so}\:{b}=\mathrm{2}{rcos}^{\mathrm{2}} \overset{} {\theta}={r}\left(\mathrm{1}+{cos}\mathrm{2}\theta\right) \\ $$$${area}\:{of}\:{triangle}=\frac{\mathrm{1}}{\mathrm{2}}{ab} \\ $$$$\:{s}\:\:=\frac{\mathrm{1}}{\mathrm{2}}{rsin}\mathrm{2}\theta.{r}\left(\mathrm{1}+{cos}\mathrm{2}\theta\right) \\ $$$$\:\:{s}=\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left({sin}\mathrm{2}\theta\right)\left(\mathrm{1}+{cos}\mathrm{2}\theta\right) \\ $$$$\frac{{ds}}{{d}\theta}=\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left\{{sin}\mathrm{2}\theta.−\mathrm{2}{sin}\mathrm{2}\theta+\left(\mathrm{1}+{cos}\mathrm{2}\theta\right)\left({cos}\mathrm{2}\theta\right).\mathrm{2}\right\} \\ $$$$\frac{{ds}}{{d}\theta}=\mathrm{0} \\ $$$$−\mathrm{2}{sin}^{\mathrm{2}} \mathrm{2}\theta+\mathrm{2}{cos}\mathrm{2}\theta+\mathrm{2}{cos}^{\mathrm{2}} \mathrm{2}\theta=\mathrm{0} \\ $$$${cos}\mathrm{4}\theta+{cos}\mathrm{2}\theta=\mathrm{0} \\ $$$$\mathrm{2}{cos}\mathrm{3}\theta.{cos}\theta=\mathrm{0} \\ $$$${if}\:{cos}\theta=\mathrm{0}\:\:{so}\:\theta=\mathrm{90}^{{o}} \:{whichis}\:{not}\:{feasible} \\ $$$${so}\:{cos}\mathrm{3}\theta=\mathrm{0}={cos}\mathrm{90}^{{o}} \\ $$$$\theta=\mathrm{30}^{{o}} \:\:{area}\:{max}=\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left({sin}\mathrm{60}^{{o}} \right)\left(\mathrm{1}+{cos}\mathrm{60}^{{o}} \right) \\ $$$$=\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}}{r}^{\mathrm{2}} \\ $$

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