The abscissa of the point on the curve 9y2 = x3 , the normal at which cuts off equal intercepts on the coordinate axes is

# The abscissa of the point on the curve , the normal at which cuts off equal intercepts on the coordinate axes is

1. A

2

2. B

4

3. C

-4

4. D

-2

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### Solution:

Differentiating $9{y}^{2}={x}^{3}$ we have $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{{x}^{2}}{6y}$

Any point on the curve is of the form $\left({t}^{2},{t}^{3}/3\right)$ and so $\frac{\mathrm{d}y}{\mathrm{d}x}$ at this point is $t/2$ Thus an equation of normal is $y-{t}^{3}/3=\left(-2/t\right)\left(X-{t}^{2}\right)$  This will intersect coordinate axes at $\left(0,2t+{t}^{3}/3\right)$ and $\left(\frac{{t}^{4}}{6}+{t}^{2},0\right)$

Hence we must have  $2t+\frac{{t}^{3}}{3}={t}^{2}+\frac{{t}^{4}}{6}⇒2+\frac{{t}^{2}}{3}=t+\frac{{t}^{3}}{6}$

Clearly  satisfies, the last equation. Hence the abscissa of the required point is 4. (For the normal meets both the axes only at origin.)  