At what point on the curve `x^(3) - 8a^(2) y= 0` the slope of the normal is `-2//3` ?

The point on the curve y=2x^(2) , where the slope of the tangent is 8, is :

At what point on the curve y = 2x^2 - 4x + 3 , the normal is parallel to Y-axis.

If a line is tangent to one point and normal at another point on the curve x=4t^(2)+3,y=8t^(3)-1, then slope of such a line is

(i) If the tangent at any point p(4m^(2), 8m^(3)) of x^(3) - y^(2)=0 is a normal to the curve x^(3) - y^(2) = 0 , then find the value of 9m^(2) .

If a line is tangent to one point and normal at another point on the curve x=4t^2+3,y=8t^3-1, then slope of such a line is

Find the equation of the tangent to the curve x = y^2-1 at the point where the slope of the normal to the curve is 2.

The slope of normal to the curve x^(3)=8a^(2)y, a gt 0 at a point in the first quadrant is -(2)/(3) , then point is

If the slope of the normal to the curve x^(3)=8a^(2)y, a gt 0 at a point in the first quadrant is -(2)/(3) , then the point is :