Question

Let the abscissa of the point on the curve $a{y}^2=x^3,$ the normal at which cuts off equal intercepts from the axes be ka/m. Find $m-k$?

Answer 1

$a{y}^2=x^3$ ...(1)
$\therefore 2ay\frac{dy}{dx}=3x^2$ or $\frac{dy}{dx}=\frac{3x^2}{2ay}$
Hence slope of the normal is $-\frac{1}{dy/dx}=\frac{-2ay}{3x^2}$
Since the normal makes equal intercepts on the axes its inclination to axis of x either ${45}^{\circ }$ or ${135}^{\circ }.$
In other words its slope is $\tan {45}^{\circ }$ or $\tan {135}^{\circ }.$ or $1,-1$.
$\therefore -\frac{2ay}{3x^2}=\pm 1$ or $4a^2{y}^2=9x^4$
$\Rightarrow 4a\cdot x^3=9x^4\therefore x=4a/9$