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

The normal lines to a given curve at each point pass through (2,0)dot The curve pass through (2,3) . Formulate the differntioal equation and hence find out the equation of the curve.

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)

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) .

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) .

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 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

Find the points on the curve y=x^3 where the slope of the tangent is equal to x-coordinate of the point.

Find the point on the curve y=x^2 where the slope of the tangent is equal to the x-coordinate of the point.

Find the points on the curve y=x^3 at which the slope of the tangent is equal to the y-coordinate of the point.

Find the points on the curve y=x^3 at which the slope of the tangent is equal to the y\ coordinate of the point.