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 perpendicular from the origin to the tangent at any point on a curve is equal to the abscissa of the point of contact.Also curve passes through the point (1,1). Then the length of intercept of the curve on the x-axis is

The perpendicular from the origin to the tangent at any point on a curve is equal to the abscissa of the point of contact. Find the equation of the curve satisfying the above condition and which passes through (1, 1).

It x intercept of any tangent is 3 xx the x- coordinate of the point of tangent,then the equation of the curve,given that it passes through (1,1) 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 .

The slope of the tangent to the curve at any point is equal to y + 2x. Find the equation of the curve passing through the origin .

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

IF slope of tangent to curve y=x^3 at a point is equal to ordinate of point , then point is

The slope of the tangent to the curve at any point is twice the ordinate at that point.The curve passes through the point (4;3). Determine its equation.