Ellipoint! #2

If we substitute this given point into the ellipse, we obtain: $\frac{h^2}{6} + \frac{1^2}{3} = 1 \Rightarrow \boxed{ h = \pm 2}.$ Going forward, let's utilize the point $(2,1)$ and examine its normal line. Taking the derivative at this point produces the slope of the tangent line:

$y = \sqrt{3 - \frac{x^2}{2}} \Rightarrow \frac{dy}{dx}|_{x=2} = \frac{-x}{2 \cdot \sqrt{3 - x^2 / 2}} \Rightarrow \frac{-2}{2 \cdot \sqrt{3 - 2^2 / 2}} = -1$

which in turn the slope of normal line through $(2,1)$ equals 1 (which is normal to the line $y = -x + 8).$