From any point on the line  $(t+2)(x+y)=1,t\ne -2$, tangents are drawn to the ellipse $4x^2+16y^2=1$. It is given that chord of contact passes through a fixed point. Then number of integral value(s) of ‘t’ for which the fixed point always lies inside the ellipse is

Any point on  $(t+2)(x+y)=1\text{\hspace{0.17em}is\hspace{0.17em}}(\alpha ,\frac{1}{t+2}-\alpha )$
Equation of chord of contact is  $4x\alpha +16y$
 $(\frac{1}{t+2}-\alpha )=1$
 $\Rightarrow \alpha (4x-16y)+(\frac{16}{t+2}y-1)=0$
The chord of contact passes through the intersection of $x-4y=0$ and  $(\frac{16}{t+2}y-1)=0$
$\therefore$ The point of intersection is  $(\frac{t+2}{4},\frac{t+2}{16})$
It lies inside the ellipse if $4{\left(\frac{t+2}{4}\right)}^2+16{\left(\frac{t+2}{16}\right)}^2-1<0$

$=(-\frac{4}{\sqrt{5}}-2)<t<(\frac{4}{\sqrt{5}}-2)\Rightarrow t=-3,-1$