Question-140748

Question Number 140748 by 676597498 last updated on 12/May/21

Answered by qaz last updated on 12/May/21

(i)...r=(2/(1+sin θ)) ⇒r+rsin θ=2 ⇒(√(x^2 +y^2 ))+y=2 ⇒x^2 +4y−4=0 −−−−−−−−−− (ii)...((4/3),(π/6))⇒((4/3)cos (π/6),(4/3)sin (π/6))=((2/( (√3))),(2/3)) x^2 +4y−4=0 ⇒y((2/( (√3))))′=−(1/( (√3))) ⇒y−(2/3)=−(1/( (√3)))(x−(2/( (√3)))) ⇒y=−(x/( (√3)))+(4/3) ⇒rsin θ=−(1/( (√3)))rcos θ+(4/3) ⇒r=(4/(3sin θ+(√3)cos θ))

$$\left({i}\right)…{r}=\frac{\mathrm{2}}{\mathrm{1}+\mathrm{sin}\:\theta} \\ $$$$\Rightarrow{r}+{r}\mathrm{sin}\:\theta=\mathrm{2} \\ $$$$\Rightarrow\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }+{y}=\mathrm{2} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +\mathrm{4}{y}−\mathrm{4}=\mathrm{0} \\ $$$$−−−−−−−−−− \\ $$$$\left({ii}\right)…\left(\frac{\mathrm{4}}{\mathrm{3}},\frac{\pi}{\mathrm{6}}\right)\Rightarrow\left(\frac{\mathrm{4}}{\mathrm{3}}\mathrm{cos}\:\frac{\pi}{\mathrm{6}},\frac{\mathrm{4}}{\mathrm{3}}\mathrm{sin}\:\frac{\pi}{\mathrm{6}}\right)=\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}},\frac{\mathrm{2}}{\mathrm{3}}\right) \\ $$$${x}^{\mathrm{2}} +\mathrm{4}{y}−\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow{y}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right)'=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{y}−\frac{\mathrm{2}}{\mathrm{3}}=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left({x}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$$\Rightarrow{y}=−\frac{{x}}{\:\sqrt{\mathrm{3}}}+\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\Rightarrow{r}\mathrm{sin}\:\theta=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}{r}\mathrm{cos}\:\theta+\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\Rightarrow{r}=\frac{\mathrm{4}}{\mathrm{3sin}\:\theta+\sqrt{\mathrm{3}}\mathrm{cos}\:\theta} \\ $$

Answered by Dwaipayan Shikari last updated on 12/May/21

r=(2/(1+sinθ)) ⇒(√(x^2 +y^2 )) +y=2 ((√(x^2 +y^2 ))=2−y) ⇒x^2 +y^2 +y^2 +2y(√(x^2 +y^2 ))=4 ⇒x^2 +y^2 +y^2 +2y(2−y)=4 ⇒4y+x^2 −4=0

$${r}=\frac{\mathrm{2}}{\mathrm{1}+{sin}\theta} \\ $$$$\Rightarrow\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:+{y}=\mathrm{2}\:\:\:\left(\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }=\mathrm{2}−{y}\right) \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{y}\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }=\mathrm{4} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{y}\left(\mathrm{2}−{y}\right)=\mathrm{4} \\ $$$$\Rightarrow\mathrm{4}{y}+{x}^{\mathrm{2}} −\mathrm{4}=\mathrm{0} \\ $$

Answered by TheSupreme last updated on 12/May/21

x=rcos(θ)=(2/(1+sin(θ)))cos(θ) y=rsin(θ)=(2/(1+sin(θ)))sin(θ) y+ysin(θ)=2sin(θ) sin(θ)(2−y)=y sin(θ)=(y/(2−y)) cos(θ)=(√(((2−y)^2 −y^2 )/((2−y)^2 )))=((√(2(2−2y)))/((2−y))) tan(θ)=(y/( (√(2(2−2y))))) x=(y/(tan(θ)))=(√(2(2−2y))) x^2 =4−4y

$${x}={rcos}\left(\theta\right)=\frac{\mathrm{2}}{\mathrm{1}+{sin}\left(\theta\right)}{cos}\left(\theta\right) \\ $$$${y}={rsin}\left(\theta\right)=\frac{\mathrm{2}}{\mathrm{1}+{sin}\left(\theta\right)}{sin}\left(\theta\right) \\ $$$${y}+{ysin}\left(\theta\right)=\mathrm{2}{sin}\left(\theta\right) \\ $$$${sin}\left(\theta\right)\left(\mathrm{2}−{y}\right)={y} \\ $$$${sin}\left(\theta\right)=\frac{{y}}{\mathrm{2}−{y}} \\ $$$${cos}\left(\theta\right)=\sqrt{\frac{\left(\mathrm{2}−{y}\right)^{\mathrm{2}} −{y}^{\mathrm{2}} }{\left(\mathrm{2}−{y}\right)^{\mathrm{2}} }}=\frac{\sqrt{\mathrm{2}\left(\mathrm{2}−\mathrm{2}{y}\right)}}{\left(\mathrm{2}−{y}\right)} \\ $$$${tan}\left(\theta\right)=\frac{{y}}{\:\sqrt{\mathrm{2}\left(\mathrm{2}−\mathrm{2}{y}\right)}} \\ $$$${x}=\frac{{y}}{{tan}\left(\theta\right)}=\sqrt{\mathrm{2}\left(\mathrm{2}−\mathrm{2}{y}\right)} \\ $$$${x}^{\mathrm{2}} =\mathrm{4}−\mathrm{4}{y} \\ $$

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