Question Number 47939 by rahul 19 last updated on 17/Nov/18 | ||
Commented by rahul 19 last updated on 17/Nov/18 | ||
$${how}\:{to}\:{draw}\:{this}\:{graph}? \\ $$$${steps}\:{plss} \\ $$ | ||
Commented by rahul 19 last updated on 17/Nov/18 | ||
$${The}\:{eq}^{{n}} \:{is}\:{satisfied}\:{by}\:{origin}\left(\mathrm{0},\mathrm{0}\right) \\ $$$${but}\:{still}\:{the}\:{graph}\:{does}\:{not}\:{pass}\:{through} \\ $$$${it}.{why}?? \\ $$ | ||
Answered by mr W last updated on 17/Nov/18 | ||
$${point}\:\left(\mathrm{0},\mathrm{0}\right)\:{is}\:{also}\:{a}\:{part}\:{of}\:{the}\:{diagram}. \\ $$$$ \\ $$$${r}^{\mathrm{4}} =\mathrm{2}{r}^{\mathrm{2}} \left(\mathrm{3}\:\mathrm{cos}^{\mathrm{2}} \:\theta+\mathrm{sin}^{\mathrm{2}} \:\theta\right)=\mathrm{2}{r}^{\mathrm{2}} \left(\mathrm{2}+\mathrm{cos}\:\mathrm{2}\theta\right) \\ $$$$\Rightarrow{r}=\mathrm{0}\:\Rightarrow{this}\:{point}\:\left(\mathrm{0},\mathrm{0}\right) \\ $$$$\Rightarrow{r}=\sqrt{\mathrm{2}\left(\mathrm{2}+\mathrm{cos}\:\mathrm{2}\theta\right)}\:\Rightarrow{this}\:{is}\:{the}\:{peanut}\:{curve} \\ $$ | ||
Commented by rahul 19 last updated on 17/Nov/18 | ||
thanks sir. | ||