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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

$${how}\:{to}\:{draw}\:{this}\:{graph}? \\ $$$${steps}\:{plss} \\ $$

Commented by rahul 19 last updated on 17/Nov/18

The eq^n  is satisfied by origin(0,0)  but still the graph does not pass through  it.why??

$${The}\:{eq}^{{n}} \:{is}\:{satisfied}\:{by}\:{origin}\left(\mathrm{0},\mathrm{0}\right) \\ $$$${but}\:{still}\:{the}\:{graph}\:{does}\:{not}\:{pass}\:{through} \\ $$$${it}.{why}?? \\ $$

Answered by mr W last updated on 17/Nov/18

point (0,0) is also a part of the diagram.    r^4 =2r^2 (3 cos^2  θ+sin^2  θ)=2r^2 (2+cos 2θ)  ⇒r=0 ⇒this point (0,0)  ⇒r=(√(2(2+cos 2θ))) ⇒this is the peanut curve

$${point}\:\left(\mathrm{0},\mathrm{0}\right)\:{is}\:{also}\:{a}\:{part}\:{of}\:{the}\:{diagram}. \\ $$$$ \\ $$$${r}^{\mathrm{4}} =\mathrm{2}{r}^{\mathrm{2}} \left(\mathrm{3}\:\mathrm{cos}^{\mathrm{2}} \:\theta+\mathrm{sin}^{\mathrm{2}} \:\theta\right)=\mathrm{2}{r}^{\mathrm{2}} \left(\mathrm{2}+\mathrm{cos}\:\mathrm{2}\theta\right) \\ $$$$\Rightarrow{r}=\mathrm{0}\:\Rightarrow{this}\:{point}\:\left(\mathrm{0},\mathrm{0}\right) \\ $$$$\Rightarrow{r}=\sqrt{\mathrm{2}\left(\mathrm{2}+\mathrm{cos}\:\mathrm{2}\theta\right)}\:\Rightarrow{this}\:{is}\:{the}\:{peanut}\:{curve} \\ $$

Commented by rahul 19 last updated on 17/Nov/18

thanks sir.

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