If  the focus of hyperbola is $(\pm 3,0)$  and the equation of a tangent is  $2x+y+4=0$, then the equation of the hyperbola is

Given  $(\pm ae,o)=(\pm 3,0)$
 $\Rightarrow ae=3\Rightarrow a^2{e}^2=9\Rightarrow b^2+a^2=\mathrm{9....}\left(i\right)$
 $\because 2x+y+4=0$
$\Rightarrow y=-2x-4$  is the tangent to the hyperbola
 $\therefore {(-4)}^2=a^2{(-2)}^2-b^2(\because c^2=a^2{m}^2-b^2)$
 $\Rightarrow 4a^2-b^2=\mathrm{16......}\left(ii\right)$
On solving Eqs.(i) and (ii), we get,
 $a^2=5,b^2=4$
$\therefore$  Equation of hyperbola is  $\frac{x^2}{5}-\frac{y^2}{4}=1$
 $\Rightarrow 4x^2-5y^2=20$