A normal at `P(x , y)` on a curve meets the x-axis at `Q` and `N` is the foot of the ordinate at `Pdot` If `N Q=(x(1+y^2))/(1+x^2)` , then the equation of curve given that it passes through the point `(3,1)` is

Find the equation of the normal to the curve x^(2) = 4y which passes through the point (1, 2).

Find the equation of the curve whose slope is (y-1)/(x^(2)+x) and which passes through the point (1,0).

At any point (x,y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (-4,-3) . Find the equation of the curve, given that it passes through (-2,1) .

For the differential equation xy(dy)/(dx) = (x + 2)( y + 2) , find the solution curve passing through the point (1, -1).

A normal is drawn at a point P(x , y) of a curve. It meets the x-axis and the y-axis in point A AND B , respectively, such that 1/(O A)+1/(O B) =1, where O is the origin. Find the equation of such a curve passing through (5, 4)

A curve passes through the point (1,pi/6) . Let the slope of the curve at each point (x , y) be y/x+sec(y/x),x > 0. Then the equation of the curve is

The equation of the curves through the point (1, 0) and whose slope is (y-1)/(x^2+x) is

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