Help pls for a curve equation(paremetric)?

Question

The equation of a curve is x^3 + 2y = 3xy

dy/dx = (y -x^2) / ( 2y^2 - x)

^ means to the power of.

Find the coordinates of the point, other than the origin, where the curve has a tangent which is parallel to the x-axis.

Thanks a lot in advance

Answer 1

x^3 + 2y = 3xy
3x^2 + 2(dy/dx) = 3y + 3x(dy/dx)
dy/dx = 3(x^2 - y)/(3x - 2)

Thus the slope of the curve at the point (h,k) is 3(h^2 - k)/(3h - 2).

If the curve has got to have a tangent parallel to x-axis i.e. with slope 0, then
3(h^2 - k)/(3h - 2) = 0
k = h^2

So at all the points with coordinates of the form (h, h^2) the tangent to the curve is parallel to x-axis.

The examples of those points are: (-1,1), (1,1), (-2,4), (2,4), etc.

However, (2/3, 4/9) is excluded from the possible list of coordinates where the tangent is parallel to x-axis as at the said point the slope is infinity i.e. parallel to y-axis.

Answer 2

I don't know where you got that derivative. Using implicit differentiation, we get 3x^2 + 2dy/dx = 3x(dy/dx) + 3y, so
dy/dx = (3y - 3x^2)/(2 - 3x). Therefore, we have slope 0 when 3y - 3x^2 = 0, or y = x^2.

Using y = x^2 in the equation for the curve, we find x^3 + 2x^2 = 3x(x^2), or 0 = 2(x^2)*(x - 1). So at x = 1, the tangent is horizontal, and y = x^2 = 1. Thus, at (1,1) we have a horizontal tangent.