The line `y=x+1` is a tangent to the curve `y^(2)=4x` at the point :

The line y=x+1 is tangent to the curve y^(2)=4x at the point :

The line y=x+1 is a tangent to the curve y^2=4 x at the point a) (1,2) b) (2,1) c) (1,-2) d) (-1,2)

The line y = x + 2 , is a tangent to the curve y^2 = 4x at the point :

The line y=m x+1 is a tangent to the curve y^2=4x Find the value of m .

The slope of the tangent to the curve y=x^(2) -x at the point where the line y = 2 cuts the curve in the first quadrant, is

The slope of the tangent to the curve y=x^(2) -x at the point where the line y = 2 cuts the curve in the first quadrant, is

A straight line parallel to the line 2x-y+5 =0 is also a tangent to the curve y^2=4x+5 . Then the point of contact is

Find the slope of tangent of the curve y =4x at the point (-1,4).