The slope of the normal to the curve $x^3+y^3=8xy$  at the point, other then origin, where it meets the curve $y^2=4x$  is__________.

Then curves are $x^3+y^3=8xy$ …..(1)  and $y^2=4x$ …..(2)

Solving $\left(1\right)$  and $\left(2\right),$  we get  $x^3+y^3=8xy$

                      $\Rightarrow x^3=4xy$

                      $\Rightarrow x^3=4x2\sqrt{x}$

                      $\Rightarrow x^{3/2}\left(x^{3/2}-8\right)=0$

                       $\Rightarrow x=0orx=2^2=4$

Now when $x=0,$  we get $y=0$

And when $x=4,$ we get $y^2=160ry\pm 4.$

But $x=4$  and $y=-4$  do not satisfy $\left(1\right).$

Thus $\left(0,0\right)$  and $\left(4,4\right)$  are the points of intersection of$\left(1\right)$  and  $\left(2\right)$

Differentiating $\left(1\right)$ , we get

$\frac{dy}{dx}=\frac{8y-3x^2}{3y^2-8x}.$

at$\left(4,4\right),\frac{dy}{dx}=-1.$

Hence, the equation of the normal to the curve $\left(1\right)$  at

$\left(4,4\right)\text{is}\left(y-4\right)=1\left(x-4\right)\text{or}y-x=0.$