To find the abscissa of the point on the curve where the normal cuts off equal intercepts on the coordinate axes, we can start by finding the equation of the normal at that point.

First, let's find the derivative of the curve equation with respect to x:
dy/dx = (1/3)x^(3-1)/(2*9y^(2-1))
dy/dx = (1/3)x^2/(18y)
dy/dx = x^2/(54y)

We know that the slope of the normal is the negative reciprocal of the slope of the curve at that point. So the slope of the normal is:
m = -54y/x^2

Let's assume the point on the curve where the normal cuts off equal intercepts on the coordinate axes is (a,b). This means the normal intersects the x-axis at (a,0) and the y-axis at (0,b).

Using the point-slope form of a line, the equation of the normal passing through (a,b) is:
y - b = -54b/a^2 * (x - a)

Since the normal cuts off equal intercepts on the x-axis and y-axis, the x-intercept and y-intercept are equal. Therefore, (a,0) and (0,b) are equidistant from the origin.

Using the distance formula, we can find the distance from the origin to (a,0) and (0,b):
Distance to (a,0) = sqrt(a^2 + 0^2) = a
Distance to (0,b) = sqrt(0^2 + b^2) = b

Since these distances are equal, we have:
a = b

Now, substitute a = b into the equation of the normal:
y - a = -54a/a^2 * (x - a)
y - a = -54/a * (x - a)
y = -54/a * x + 55a

Now, substitute y = 0 to find the x-intercept:
0 = -54/a * x + 55a
54/a * x = 55a
x = (55a^2)/54

Since the x-intercept is equal to the y-intercept, we have:
(55a^2)/54 = a

Simplifying this equation, we get:
55a^2 = 54a
a(55a - 54) = 0

This equation has two possible solutions:
a = 0
55a - 54 = 0
55a = 54
a = 54/55

Since a cannot be zero (as it would not define a point on the curve), the only solution is:
a = 54/55

Therefore, the abscissa of the point on the curve where the normal cuts off equal intercepts on the coordinate axes is 54/55.

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