Let R be a rectangle given by the lines $$x=0, x=2, y=0$$ and $$y=5$$. Let A$$(\alpha,0)$$ and B$$(0,\beta),\alpha\in[0,2]$$ and $$\beta\in[0,5]$$, be such that the line segment AB divides the area of the rectangle R in the ratio 4 : 1. Then, the mid-point of AB lies on a :

Let $$\mathrm{P}\left(x_{0}, y_{0}\right)$$ be the point on the hyperbola $$3 x^{2}-4 y^{2}=36$$, which is nearest to the line $$3 x+2 y=1$$. Then $$\sqrt{2}\left(y_{0}-x_{0}\right)$$ is equal to :

Let T and C respectively be the transverse and conjugate axes of the hyperbola $$16{x^2} - {y^2} + 64x + 4y + 44 = 0$$. Then the area of the region above the parabola $${x^2} = y + 4$$, below the transverse axis T and on the right of the conjugate axis C is :