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

A curve y=f(x) passes through the 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. Determine the equation of the curve. Also obtain the area bounded by the y-axis, the curve and the normal to the curve at P .

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

The slope of tangent at a point P(x, y) on a curve is - x/y . If the curve passes through the point (3, -4) , find the equation of the curve.

At any point on the curve y=f(x) ,the sub-tangent, the ordinate of the point and the sub-normal are in

The slope of the tangent at a point P(x, y) on a curve is (- (y+3)/(x+2)) . If the curve passes through the origin, find the equation of the curve.

The x-intercept of the tangent to a curve is equal to the ordinate of the point of contact. The equation of the curve through the point (1,1) is

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

If length of tangent at any point on th curve y=f(x) intercepted between the point and the x -axis is of length 1. Find the equation of the curve.

The slope of a curve at any point is the reciprocal of twice the ordinate at that point and it passes through the point (4,3). The equation of the curve is: