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Hola alexander perez!
Te hago la primera. Manda la otra en otra pregunta(se colapsa el editor de ecuaciones)
$$\begin{align}&r=\sqrt {x^2+y^2}\\&\\&\cos \theta=\frac{r}{\sqrt{x^2+y^2}}\\&\\&r=\frac{2}{3-2 \cos \theta}\\&\\&\sqrt {x^2+y^2}=\frac{2}{3-2 \frac{x}{\sqrt {x^2+y^2}}}\\&simplificando \ el \ radical:\\&\\&\sqrt {x^2+y^2}=\frac{2 \sqrt {x^2+y^2}}{3 \sqrt {x^2+y^2}-2x}\\&\\&3 \sqrt {x^2+y^2}-2x=2\\&\\&3 \sqrt {x^2+y^2} =2+2x\\&\\&\Bigg(3 \sqrt {x^2+y^2}\Bigg)^2= \big(2+2x \Big)^2\\&\\&9(x^2+y^2)=4+4x^2+8x\\&\\&5x^2-8x+9y^2=4\\&\\&completando \ cuadrados º (y º compensando \ término \ independiente):\\&\\&5 \Big(x^2-\frac{8}{5} \Big)+9y^2=4\\&\\&5 \Big(x- \frac{4}{5} \Big)^2- \frac{16}{5}+9y^2=4\\&\\&\\&\\&5 \Big(x- \frac{4}{5} \Big)^2+9y^2=\frac{36}{5}\\&\\&dividiendo \ por \ 36/5\\&\\&\frac{5 \Big(x- \frac{4}{5} \Big)^2}{\frac{36}{5}}+ \frac{9y^2}{\frac{36}{5}}=1\\&\\&\frac{\Big(x- \frac{4}{5} \Big)^2}{\frac{36}{25}}+ \frac{y^2}{\frac{4}{5}}=1\\&\\&\\&\\&\end{align}$$
es una elipse centrada en (4/5,0)
Saludos
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