Sloppy Ellipse

We can scale the ellipse horizontally by $\frac{3}{4}$ to make it into a circle of radius 3, then draw the tangent through (0,5), with slope $\frac{4}{3}$ , and scale back to get a tangent of slope $\boxed{1}$ .

Let the slope be equal to $m$ . Then, we get two equations,

$y=mx+5$

$9x^2+16y^2=144$

Now, substituting the first equation into the second, we get

$9x^2+16(mx+5)^2=144$ , $(9+16m^2)x^2+(160m)x +256$

Now, this equation will have real solutions if the discriminant is greater than or equal to zero, and if the line and the ellipse share a common point. So, we find the value of the discriminant, and make it greater than or equal to zero, giving us

$(160m)^2-(4)(9+16m^2)(256) \geq 0$ , Dividing by $256 \cdot 4$ gives us

$25m^2-9-16m^2 \geq 0$

$m^2 \geq 1$

Which gives us the smallest positive solution $\boxed{1}$