Find the abscissa of the point on the curve ay2 = x3, the normal at which cuts of equal intercept from the axes.
x = 4a/9
The given curve is ay2 = x3 …(i)
Differentiate to get:
The slope of normal
Since the normal makes equal intercepts on the axes its inclinations to axis of x is either 45o or 135o.
So two normal are possible with slopes 1 and –1.
One squaring 4a2y2 = 9x4. Using (i) we get:
4a x3 = 9x4 ⇒ x = 4a/9.
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