Solution
= Solution
By <implicit differentiation> of the <ellipse> equation,
$$
18x+32y\frac{dy}{dx}=0\quad\Longrightarrow\quad\left.\frac{dy}{dx}\right|_{(1,1)}=-\frac9{16}.
$$
The <tangent line> through $(1,1)$ is therefore
$$
\boxed{y-1=-\frac9{16}(x-1),\qquad\text{equivalently }9x+16y=25.}
$$
Its <normal vector> is $(9,16)$, parallel to the <gradient> $(18,32)$ of $9x^2+16y^2$ at that point.