If the tangent to the conic $y-6=x^2\text{ at }(2,10)$ touches the circle $x^2+y^2+8x-2y=k$  (for some fixed $k$) at a point $(\alpha ,\beta ),\text{ then }(\alpha ,\beta )$ is 

lf a line is tangent to the curve at $\left(x_1,y_1\right)$ then equation of tangent is $\left(y-y_1\right)=\frac{dy}{dx}\left(x-x_1\right)$

Given curves are $y-6=x^2\to \left(1\right),x^2+y^2+8x-2y=k\to \left(2\right)$

From Equation (1 ), $y^{\mathrm{\prime }}=2x\Rightarrow y^{\mathrm{\prime }}=4\text{ at }(2,10)$

Now, equation of tangent is $y-10=4(x-2)\Rightarrow 4x-y+2=0$  

$\because 4x-y+2=0$ passes through $(\alpha ,\beta )\Rightarrow 4\alpha -\beta +2=0\Rightarrow \beta =4\alpha +2\to \left(3\right)$

From Equation (2), we get $2x+2y{y}^{\mathrm{\prime }}+8-2y^{\mathrm{\prime }}=0\Rightarrow y^{\mathrm{\prime }}=\frac{2x+8}{2-2y}\Rightarrow y^{\mathrm{\prime }}=\frac{2\alpha +8}{2-2\beta }=4$

  $\Rightarrow 2\alpha +8=8-8\beta \Rightarrow \beta =-\frac{\alpha }{4}\to \left(4\right)$

On solving equations (3) and (4), we get  $\alpha =\frac{-8}{17},\beta =\frac{2}{17}$.