Answer:
13. [tex]x^2+y^2+2x-3\sqrt{x^2+y^2}=0[/tex]
14. [tex]r=3\cos(\theta)-2\sin(\theta)[/tex]
Step-by-step explanation:
Question 13
Conversion from Polar equation to rectangular equation:
[tex]x=r \cos(\theta) \implies \cos(\theta)=\dfrac{x}{r}[/tex]
[tex]y=r \sin(\theta) \implies \sin(\theta)=\dfrac{y}{r}[/tex]
[tex]x^2+y^2=r^2 \implies r=\sqrt{x^2+y^2}[/tex]
Given:
[tex]r=3-2\cos(\theta)[/tex]
[tex]\textsf{Substitute }\cos(\theta)=\dfrac{x}{r}:[/tex]
[tex]\implies r=3-\dfrac{2x}{r}[/tex]
Multiply both sides by r:
[tex]\implies r^2=3r-2x[/tex]
[tex]\implies r^2+2x-3r=0[/tex]
[tex]\textsf{Substitute }x^2+y^2=r^2\:\textsf{and }r=\sqrt{x^2+y^2}:[/tex]
[tex]\implies x^2+y^2+2x-3\sqrt{x^2+y^2}=0[/tex]
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Question 14
Conversion from Rectangular equation to polar equation:
[tex]x=r \cos(\theta)[/tex]
[tex]y=r \sin(\theta)[/tex]
[tex]x^2+y^2=r^2 \implies r^2\cos^2(\theta)+r^2\sin^2(\theta)=r^2[/tex]
Given:
[tex]x^2+y^2-3x+2y=0[/tex]
[tex]\textsf{Substitute }x^2+y^2=r^2\:\textsf{and }x=r \cos(\theta)\:\textsf{and }y=r \sin(\theta):[/tex]
[tex]\implies r^2-3r\cos(\theta)+2r\sin(\theta)=0[/tex]
Factor out common term r:
[tex]\implies r(r-3\cos(\theta)+2\sin(\theta))=0[/tex]
Divide both sides by r:
[tex]\implies r-3\cos(\theta)+2\sin(\theta)=0[/tex]
Rewrite to make r the subject:
- [tex]\implies r=3\cos(\theta)-2\sin(\theta)[/tex]
