The tangents to a curve at a point on it is perpendicular to the line joining the point with the origin. Find the equation of the curve.

A curve passes through the point (-2, 1) and at any point (x, y) of the 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.

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

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 Find the equation of the curve given that it passes through quad (-2,1)

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).

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.

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

The slope of the tangent at any point on a curve is lambda times the slope of the line joining the point of contact to the origin. Formulate the differential equation and hence find the equation of the curve.

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.

Find the equation of the perpendicular bisector of the line segment joining the origin and the point (4, 6) .

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