$\text{ The equation of the parabola to which the line }{m}^2(y-10)-mx-1=0\text{ is a tangent for any }$$\text{ real value of }m,\text{ is }$

Equation of the given line is

$x=m\left(y-10\right)-\frac{1}{m}$ 

Comparing with $x-h=m\left(y-k\right)+\frac{a}{m}$

 We get  $h=0,k=10\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}}a=-1$

 The equation of parabola is${\left(x-h\right)}^2=4a\left(y-k\right)$

$\Rightarrow x^2=-4\left(y-10\right)$