Question

Choose the correct answer in the following question:
The point on the curve $9y^2=x^3$, where the normal to the curve makes equal intercepts with the axes is

(a) $(4,\pm \frac{8}{3})$ (b) $(4,\frac{-8}{3})$
(c) $(4,\pm \frac{3}{8})$ (d) $(\pm 4,\frac{3}{8})$

Answer 1

Given, curve is $9y^2=x^3$ ...(i)
On differentiating, we get $18y\frac{dy}{dx}=3x^2\Rightarrow \frac{dy}{dx}=\frac{x^2}{6y}$ ...(ii)
Let $(\alpha ,\beta )$ be a point on Eq. (i) at which normal makes equal intercepts on the axes, then
$9{\beta }^2={\alpha }^3$ ...(iii)
From Eq. (ii), slope of the normal at $(\alpha ,\beta )={\frac{-1}{\left(\frac{dy}{dx}\right)}}_{(\alpha ,\beta )}=\frac{-1}{\frac{{\alpha }^2}{\left(6\beta \right)}}=\frac{-6\beta }{{\alpha }^2}$ ...(iv)
Since, normal of the curve makes equal intercepts with the axes, so slope of normal $=tan{45}^{\circ }ortan{135}^{\circ }=\pm 1$ ...(v)
$\therefore$ From Eq. (iv), we get $\frac{-6\beta }{{\alpha }^2}=\pm 1\Rightarrow \beta =\mp \frac{{\alpha }^2}{6}$
On putting the value of $\beta$ in Eq. (iii), we get
$9(\mp \frac{{\alpha }^2}{6}{)}^2={\alpha }^3\Rightarrow {\alpha }^4=4{\alpha }^3\Rightarrow {\alpha }^3(\alpha -4)=0\Rightarrow \alpha =0or\alpha =4$
When $\alpha =0,\beta =0,$ then normal passes through (0, 0) it mean that they do not intercepts.
Taking $\alpha =4,Weget\beta =\mp \frac{4^2}{6}=\mp \frac{8}{3}.$ Hence, the correct option is (a).