Question

The points on the curve $9y^2=x^3$, where the normal to the curve makes equal intercepts with the axes are

• A. $(4,\pm \frac{8}{3})$
• B. $(4,\frac{-8}{3})$
• C. $(4,\pm \frac{3}{8})$
• D. $(\pm 4,\frac{3}{8})$
• E. answer required

Answer 1

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

Let the point on curve be $(h,k)$
$y^2=\frac{x^3}{9}2y.\frac{dy}{dx}=\frac{1}{9}.3x^2\therefore \frac{dy}{dx}=\frac{x^2}{6y}$
$\frac{dy}{dx}$ gives slope of curve at given point
$\frac{dy}{dx}at(h,k)=\frac{h^2}{6k}$
but this is the slope of tangent and we need slope of normal
As we know that slope of normal$\times$ slope of tangent is $-1$
$\therefore$ slope of normal is $\frac{-6k}{h^2}$
Given in question that normal makes equal intercepts on axes which means slope of normal is either $1$ or $-1$
$-6k=h^2$ and $6k=h^2$
Now go through the option you will that option A satisfy our equation