The curve with the property that the projection of the ordinate on the normal is constant and has a length equal to `a` is

Find the curve for which the perpendicular from the foot of the ordinate to the tangent is of constant length.

A curve y=f(x) has the property that the perpendicular distance of the origin from the normal at any point P of the curve is equal to the distance of the point P from the x-axis. Then the differential equation of the curve

The Curve possessing the property that the intercept made by the tangent at any point of the curve on they-axis is equal to square of the abscissa of the point of tangency, is given by

A curve C has the property that if the tangent drawn at any point P on C meets the co-ordinate axis at A and B , then P is the mid-point of A Bdot The curve passes through the point (1,1). Determine the equation of the curve.

The equation of the curve lying in the first quadrant, such that the portion of the x - axis cut - off between the origin and the tangent at any point P is equal to the ordinate of P, is (where, c is an arbitrary constant)

A curve is such that the area of the region bounded by the co-ordinate axes, the curve & the coordinate of any point on it is equal to the cube of that ordinate. The curve represents

A curve passing through the point (1,2) and satisfying the condition that slope of the normal at any point is equal to the ratio of ordinate and abscissa of that point , then the curve also passes through the point

The slope of the tangent at any arbitrary point of a curve is twice the product of the abscissa and square of the ordinate of the point. Then, the equation of the curve is (where c is an arbitrary constant)

The equation of the tangent to the curve y=sqrt(9-2x^(2)) at the point where the ordinate & the abscissa are equal is

The curve for which the slope of the tangent at any point is equal to the ration of the abcissa to the ordinate of the point is