Respuesta :

caylus
Hello,

[tex]\left{x = 4 \sin^2 \theta , \quad y = 3 \cos^2 \theta}\\\\ sin^2(\theta)= \dfrac{4}{x} , \quad cos^2(\theta)= \dfrac{3}{y} \\\\ sin^2(\theta)+cos^2(\theta)= \dfrac{4}{x} + \dfrac{3}{y} \\\\ \dfrac{3}{y}=1-\dfrac{4}{x}\\\\ y=\dfrac{3x}{x-4} [/tex]