Question

The point on the curve 9y² = x³, where the normal to the curve makes equal intercepts with the axes is
(a) 4,$\frac{8}{3}$
(b) -4, $\frac{8}{3}$
(c) 4, $\frac{-8}{3}$
(d) none of these

Answer 1

Given curve

$9y^2=x^{3}y=\frac{x^{3/2}}{3}$

Now, diff. w. r. t $x$

$\frac{dy}{dx}=\frac{3}{2}\frac{x^{1/2}}{x^3}=\frac{1}{2}\sqrt{x}$

Now, Slope of the normal $=\frac{-2}{\sqrt{x}}=-1\Rightarrow x=4y=\sqrt{\frac{64}{9}}=\pm \frac{8}{3}$

Hence, points are $(4,8/3),(4,-8/3)$

Hence, option (a) and (c) are correct.