If `Ax+By=1` is a normal to the curve `ay=x^(2)`, then :

Find the equation of the normal to the curve ay^(2)=x^(3) at the point (am^(2), am^(3)) .

Find the equation of the normal to the curve ay^(2)=x^(3), at the point (am^(2),am^(3)). If normal passes through the point (a,0) then find the slope.

If the normal to the curve ay^(2)=x^(3)(a ne0) at a point on it makes equl intercepts with the coordinates axes, find the x-coordinates of the point.

If the equation of the normal to the curve y=ax^(2)+2x-3, at =1, is x +6y=7 then : a =

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

Find the equation of the normal to the curve ay^2 =x^3 at (am^2 ,am^3)

The value of the parameter a so that the line (3-a)x+ay+ a^2-1=0 is normal to the curve xy=1 may lie in the interval

The value of the parameter a so that the line (3-a)x+ay+a^(2)-1=0 is normal to the curve xy=1 may lie in the interval