Question

Find the point on the curve $9y^2=x^3$ while normal to the curve makes equal intercepts with the axes

• A. $(4,\frac{8}{3})$
• B. $(4,\frac{5}{3})$
• C. $(2,\frac{8}{3})$
• D. $(4,\frac{7}{3})$

Answer 1

The correct option is A $(4,\frac{8}{3})$

Let ($x_1,y_1$) be a point on the curve $9y^2=x^3$ where normal to the curve makes equal intercepts with the axes.

Differentiating w.r.t x.

$9\left(2y\right)\frac{dy}{dr}=3x^2\Rightarrow \frac{dy}{dx}=\frac{x^2}{6y}$

The slope of the normal to the given curve at point ($x_1,y_1$)

$=\frac{-1}{\frac{dy}{dx}(x_1,y_1)}=\frac{-6y_1}{x_1^2}$

$\therefore$ The equation of the normal to the curve at ($x_1,y_1$) is

$y-y_1=\frac{-6y_1}{x_1^2}(x-x_1)\because [y-y_1=m(x-x_1]\Rightarrow x_1^{2}y-x_1^2{y}_1=-6x{y}_1+6x_1{y}_1\Rightarrow \frac{\theta x{y}_1}{6x_1{y}_1+x_1^2{y}_1}+\frac{x_1^{2}y}{6x_1{y}_1+x_1^2{y}_1}=1\therefore \frac{x}{\frac{x_1(6+x_1)}{6}}+\frac{y}{\frac{y_1(6+x_1)}{x_1}}=1$

The normal makes equal intercepts with the axes.

$\therefore \frac{x_1(6+x_1)}{6}=\frac{y_1(6+x_1)}{x_1}\Rightarrow \frac{x_1}{6}=\frac{y_1}{x_1}\therefore x_1^2=6y_1\to \left(1\right)$

Also the point $x_1,y_1$

Lies on the curve , $9y_1^2=x_1^3\to \left(2\right)$

$\Rightarrow 9{\left(\frac{x_1^2}{6}\right)}^2=x_1^3\Rightarrow \frac{x_1^4}{4}=x_1^3\therefore x_1=4$

From (2)$\Rightarrow 9y_1^2=4^3=64$

$\therefore y_1=\frac{8}{3}$

Therefore required point $(4,\frac{8}{3})$