Thank you for reporting, we will resolve it shortly
Q.
If the tangent to the conic, $y — 6 = x^2$ at ($2, 10$ ) touches the circle, $x^2 + y^2 + 8x — 2y = k$ (for some fixed $k$ ) at a point $\left(\alpha, \beta\right)$; then $\left(\alpha, \beta\right)$ is :
Equation tangent to $y=x^{2}+6$ at $(2,10)$ is
$\left(\frac{y+10}{2}\right)=2 x+6$ i.e. $4 x-y+2=0$...(1)
$\because(1)$ is tangent to circle $x^{2}+y^{2}+8 x-2 y=k$
Also $P(\alpha, \beta)$ must be foot of perpendicular from
$C(-4,1)$ to $4 x-y+2=0$
$\Rightarrow \frac{x+4}{4}=\frac{y-1}{-1}=\frac{[4(-4)-1+2]}{16+1}$
$\Rightarrow x+4=\frac{60}{17}, y-1=\frac{-15}{17}$
$\Rightarrow x=\frac{-8}{17}, y=\frac{2}{17}$
$\Rightarrow (\alpha, \beta) \equiv\left(\frac{-8}{17}, \frac{2}{17}\right)$
