Question

The points on the curve 9y² = x³, where the normal to the curve make equal intercepts with the axes are

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

Answer 1

(a) $\left(4,\pm \frac{8}{3}\right)$

Let (x₁, y₁) be the required point.
$$\begin{gathered} \mathrm{Since}\left(x_1,y_1\right)\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{given}\mathrm{curve} \\ \therefore 9{y_1}^2={x_1}^3...\left(1\right) \\ \mathrm{Now},9y^2=x^3 \\ 18y\frac{dy}{dx}=3x^2 \\ \Rightarrow \frac{dy}{dx}=\frac{3x^2}{18y}=\frac{x^2}{6y} \\ \text{Slope of the tangent =}{\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}\text{=}\frac{{x_1}^2}{6y_1} \\ \text{Slope of the normal =}\frac{-1}{\frac{{x_1}^2}{6y_1}}\text{=}\frac{-6y_1}{{x_1}^2} \\ \text{It is given that the normal makes equal intercepts with the axes.} \\ \text{∴ Slope of the normal = ±1} \\ \mathrm{Now}, \\ \frac{-6y_1}{{x_1}^2}=\pm 1 \\ \Rightarrow \frac{-6y_1}{{x_1}^2}=1\text{or}\frac{-6y_1}{{x_1}^2}\text{=-1} \\ \Rightarrow y_1=\frac{-{x_1}^2}{6}\text{or}{y}_1=\frac{{x_1}^2}{6}...\left(2\right) \\ \text{Case 1: When}{y}_1=\frac{-{x_1}^2}{6} \\ \text{From (1), we have} \\ 9\left(\frac{{x_1}^4}{36}\right)={x_1}^3 \\ \Rightarrow {x_1}^4=4{x_1}^3 \\ \Rightarrow {x_1}^4-4{x_1}^3=0 \\ \Rightarrow {x_1}^3\left(x_1-4\right)=0 \\ \Rightarrow x_1=0,4 \\ \mathrm{Putting}{x}_1=0\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get}, \\ 9{y_1}^2=0 \\ \Rightarrow y_1=0 \\ \mathrm{Putting}{x}_1=4\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get}, \\ 9{y_1}^2=4^3 \\ \Rightarrow y_1=\pm \frac{8}{3} \\ \mathrm{But},\mathrm{the}\mathrm{line}\mathrm{making}\mathrm{the}\mathrm{equal}\mathrm{intercepts}\mathrm{with}\mathrm{the}\mathrm{coordinate}\mathrm{axes}\mathrm{can}\mathrm{not}\mathrm{pass}\mathrm{through}\mathrm{the}\mathrm{origin}. \\ \mathrm{Hence},\mathrm{the}\mathrm{points}\mathrm{are}\left(4,\pm \frac{8}{3}\right). \end{gathered}$$