If the normal from the point$(h, 1)$ on the ellipse $\frac{x^2}{6}+\frac{y^2}{3}=1$ is perpendicular to the line $x + y = 8$, then the value of $h$ is

The slope of the normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

at the point $\left(x_1,y_1\right)$ is $\frac{a^2{y}_1}{b^2{x}_1}$ So, the slope of the normal to 

$\frac{x^2}{6}+\frac{y^2}{3}=1\text{ at }P(h,1)\text{ is }\frac{6}{3h}=\frac{2}{h}\text{. }$

It is given that the normal is perpendicular to the line $x + y = 8.$

$\therefore \frac{2}{h}\times -1=-1\Rightarrow h=2$