Question Number 136988 by mohammad17 last updated on 28/Mar/21
$${whats}\:{the}\:{intersiction}\:{of}\:{the}\:{curves}\:{r}=\frac{\mathrm{1}}{\mathrm{1}−{cos}\theta} \\ $$$$ \\ $$$${and}\:{r}=\frac{\mathrm{1}}{\mathrm{1}+{cos}\theta}\:? \\ $$
Answered by Ñï= last updated on 28/Mar/21
$${r}=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{cos}\:\theta} \\ $$$$\Rightarrow{r}−{r}\mathrm{cos}\:\theta=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }−{x}=\mathrm{1}\:\:\:\left(\mathrm{1}\right) \\ $$$${r}=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{cos}\:\theta} \\ $$$$\Rightarrow{r}+{r}\mathrm{cos}\:\theta=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }+{x}=\mathrm{1}\:\:\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{2}\right)\wedge\left(\mathrm{1}\right) \\ $$$$\Rightarrow{x}=\mathrm{0}\:\:{y}=\pm\mathrm{1} \\ $$$${intersect}\:{points}:\left(\mathrm{0},\mathrm{1}\right),\left(\mathrm{0},−\mathrm{1}\right) \\ $$
Answered by physicstutes last updated on 29/Mar/21
$$\mathrm{At}\:\mathrm{the}\:\mathrm{point}\:\mathrm{of}\:\mathrm{intersection},\mathrm{these}\:\mathrm{two} \\ $$$$\mathrm{polar}\:\mathrm{curves}\:\mathrm{are}\:\mathrm{equal}.\:\mathrm{That}\:\mathrm{is} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{1}−\:\mathrm{cos}\:\theta}\:=\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{cos}\:\theta} \\ $$$$\Rightarrow\:\mathrm{1}−\mathrm{cos}\:\theta\:=\:\mathrm{1}\:+\:\mathrm{cos}\:\theta \\ $$$$\mathrm{cos}\:\theta\:=\:\mathrm{0}\:\mathrm{or}\:\theta\:=\:\mathrm{2}\pi{n}\pm\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{hence}\:\theta=\left\{\frac{\pi}{\mathrm{2}},−\frac{\pi}{\mathrm{2}}\right\} \\ $$$$\mathrm{points}\:\mathrm{of}\:\mathrm{intersection}\:\mathrm{are}\:\left(\mathrm{1},\frac{\pi}{\mathrm{2}}\right)\:\mathrm{and}\:\left(−\mathrm{1},−\frac{\pi}{\mathrm{2}}\right) \\ $$$$ \\ $$
Commented by physicstutes last updated on 29/Mar/21
