The product of the slopes of the common tangents of the ellipse $x^2+4y^2=16$ and the parabola $y^2-4x-4=0$ is

Given, ellipse: $\frac{x^2}{16}+\frac{y^2}{4}=1$ and parabola: $y^2=4\left(x+1\right)$
Putting $x+1=X$ and $Y=y$, we get,

$Y^2=4X$

So, the equation of the tangent is $Y=mX+\frac{1}{m}=m\left(x+1\right)+\frac{1}{m}$
$Y=mx+\left(m+\frac{1}{m}\right)\dots \left(\mathrm{i}\right)$
If $\left(\mathrm{i}\right)$ is also a tangent to $\frac{x^2}{16}+\frac{y^2}{4}=1$,

then, $c^2=a^2{m}^2+b^2$
$\Rightarrow {\left(m+\frac{1}{m}\right)}^2=16m^2+4$
$\Rightarrow m^2+\frac{1}{m^2}+2=16m^2+4$
$\Rightarrow 15m^2-\frac{1}{m^2}+2=0$
Let, $m^2=t>0$

$\Rightarrow 15t^2+2t-1=0$

$\Rightarrow t=\frac{1}{5},\frac{-1}{3}\Rightarrow m^2=\frac{1}{5}$

Hence, $m=\pm \sqrt{\frac{1}{5}}$