MathematicsMedium57×since 2003Q3992Let P(x0,y0)\mathrm{P}\left(x_{0}, y_{0}\right)P(x0,y0) be the point on the hyperbola 3x2−4y2=363 x^{2}-4 y^{2}=363x2−4y2=36, which is nearest to the line 3x+2y=13 x+2 y=13x+2y=1. Then 2(y0−x0)\sqrt{2}\left(y_{0}-x_{0}\right)2(y0−x0) is equal to :A3B−-−9C−-−3D9Check answerSkip