The tangent at the point (2, $$-$$2) to the curve, x²y² $$-$$ 2x = 4(1 $$-$$ y) does not pass through the point :

The normal to the curve y(x – 2)(x – 3) = x + 6 at the point where the curve intersects the y-axis passes through the point :

Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the maximum area (in sq. m) of the flower-bed, is :

Let C be a curve given by y(x) = 1 + $$\sqrt {4x - 3} ,x > {3 \over 4}.$$ If P is a point on C, such that the tangent at P has slope $${2 \over 3}$$, then a point through which the normal at P passes, is :