If the tangent at any point ;$P(4m^{2}, 8m^{3})$ of $x^{3}-y^{3}=0$ is also a normal to the curve & ; $x^{3}-y^{3}=0$ , then value of m is-

To find the value of $m$, we need to use the fact that the tangent at any point $P(4m^2, 8m^3)$ on the curve $x^3 - y^3 = 0$ is also a normal to the same curve.

First, let's find the equation of the curve $x^3 - y^3 = 0$.

Given $x^3 - y^3 = 0$, we can rewrite it as $y^3 = x^3$. Taking the cube root of both sides, we get $y = x$.

So, the curve is a set of points lying on the line $y = x$.

Now, let's find the derivative of $y = x$:

$\frac{dy}{dx} = 1$

Since the slope of the tangent line is the derivative of the curve, the slope of the tangent line at any point $P$ on the curve $x^3 - y^3 = 0$ is $1$.

Now, let's find the slope of the line joining the origin $(0,0)$ and the point $P(4m^2, 8m^3)$:

$m = \frac{8m^3 - 0}{4m^2 - 0} = \frac{8m^3}{4m^2} = 2m$

We know that the slope of the tangent line is $1$, and since the tangent at $P$ is also a normal to the curve, the slope of the tangent line is perpendicular to the slope of the line joining $P$ and the origin.

So, we have:

$\text{Slope of tangent} \times \text{Slope of line joining } P \text{ and origin} = -1$

$(1) \times (2m) = -1$

$2m = -1$

$m = -\frac{1}{2}$

So, the value of $m$ is $-\frac{1}{2}$.