Respuesta :

[tex]\\ \rm\Rrightarrow r=2cos\theta-3sin\theta[/tex]

  • a(rcosTheta) +b(r sintheta)=ax+by

[tex]\\ \rm\Rrightarrow \dfrac{2}{r}cos\theta-\dfrac{3}{r} sin\theta=1[/tex]

[tex]\\ \rm\Rrightarrow 2x-3y=1[/tex]

[tex]\\ \rm\Rrightarrow 2x-3y-1=0[/tex]

semsee45

Answer:

[tex](x-1)^2+(y+1.5y)^2=3.25[/tex]

Step-by-step explanation:

[tex]x=r \cos(\theta)[/tex]

[tex]y=r \sin(\theta)[/tex]

Given polar equation:

[tex]r=2\cos(\theta)-3 \sin(\theta)[/tex]

Multiply both sides by r:

[tex]r^2=2r\cos(\theta)-3r \sin(\theta)[/tex]

Know that [tex]x^2+y^2=r^2[/tex]:

[tex]x^2+y^2=2r\cos(\theta)-3r \sin(\theta)[/tex]

Substitute [tex]x=r \cos(\theta)[/tex] and [tex]y=r \sin(\theta)[/tex] :

[tex]x^2+y^2=2x-3y[/tex]

Therefore,

[tex]x^2-2x+y^2+3y=0[/tex]

Complete the square:

[tex](x-1)^2-1+(y+1.5y)^2-2.25=0[/tex]

[tex](x-1)^2+(y+1.5y)^2=3.25[/tex]

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