The slope of the tangent at `(x , y)` to a curve passing through a point `(2,1)` is `(x^2+y^2)/(2x y)` , then the equation of the curve is

The slope of the tangent at (x , y) to a curve passing through (1,pi/4) is given by y/x-cos^2(y/x), then the equation of the curve is

A curve passes through (1,pi/4) and at (x,y) its slope is (sin 2y)/(x+tan y). Find the equation to the curve.

If slope of the tangent at the point (x, y) on the curve is (y-1)/(x^(2)+x) , then the equation of the curve passing through M(1, 0) is :

A curve y=f(x) passes through point P(1,1) . The normal to the curve at P is a (y-1)+(x-1)=0 . If the slope of the tangent at any point on the curve is proportional to the ordinate of the point, then the equation of the curve is

The slope of the tangent to the curve (y-x^5)^2=x(1+x^2)^2 at the point (1,3) is.

The slope of the tangent to a curve y=f(x) at (x,f(x)) is 2x+1. If the curve passes through the point (1,2) then the area of the region bounded by the curve, the x-axis and the line x=1 is (A) 5/6 (B) 6/5 (C) 1/6 (D) 1

The curve y=f(x) in the first quadrant is such that the y - intercept of the tangent drawn at any point P is equal to twice the ordinate of P If y=f(x) passes through Q(2, 3) , then the equation of the curve is

A curve passes through the point (3, -4) and the slope of the tangent to the curve at any point (x, y) is (-x/y) .find the equation of the curve.

Given that the slope of the tangent to a curve y = f(x) at any point (x, y ) is (2y )/ ( x ^ 2 ) . If the curve passes through the centre of the circle x ^ 2 + y ^ 2 - 2 x - 2y = 0 , then its equation is :

Find the slope of the tangent to the curve y = x^3- x at x = 2 .