The shortest distance of the point (h,k) from both the axes are
The points (3, 3), (h, 0) and (0, k) are collinear if
Without changing the direction of coordinate axes, origin is transferred to (h, k), so that the linear (one degree)
terms in the equation x2+y2−4x+6y−7=0 are eliminated. Then the point (h, k) is
Find the equation of normal to the parabola y2=4ax at point (h,k) on the parabola