Find the points on the curve `9y^2=x^3` where normal to the curve makes equal intercepts with the axes.

The points on the curve 9y^2=x^3 , where the normal to the curve makes equal intercepts with the axes are (A) (4,+-8/3) (B) (4,(-8)/3) (C) (4,+-3/8) (D) (+-4,3/8)

If (a,b) be the point on the curve 9y ^(2)=x ^(3) where normal to the curve make equal intercepts with the axis, then the value of (a+b) is:

The point on the curve 9y^2=x^3 , where the normal to the curve makes equal intercepts with the axes is (4,\ +-8//3) (b) (-4,\ 8//3) (c) (-4,\ -8//3) (d) (8//3,\ 4)

Find the point on the curve 9y^(2)=x^(3), where the normal to the curve makes equal intercepts on the axes.

(i) Find the point on the curve 9y^(2)=x^(3) where normal to the curve has non zero x -intercept and both the x intercept and y- intercept are equal. (ii) If the tangent at (1,1) on y^(2)=x(2-x)^(2) meets the curve again at P, then find coordinates of P (iii) The normal to the curve 5x^(5) -10x^(3) +x+2y+6=0 at the point P(0,-3) is tangent to the curve at some other point (s). Find those point(s)?

The abscissa of the point on the curve ay^(2)=x^(3), the normal at which cuts off equal intercepts from the coordinate axes is

The point on the curve y^(2) = 4ax at which the normal makes equal intercepts on the coordinate axes is