If the tangents on the ellipse $4x^2+y^2=8$ at the points (1,2) and (a,b) are perpendicular to each other, then $a^2=$

($a,b$) lies on $4x^2+y^2=8\Rightarrow 4a^2+b^2=8$.

Given curve is $4x^2+y^2=8\Rightarrow 8x+2y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{4x}{y}$.

${\left(\frac{dy}{dx}\right)}_{(1,2)}=\frac{-4}{2}=-2,{\left(\frac{dy}{dx}\right)}_{(a,h)}=\frac{-4a}{b}$.

Tangents at (1, 2) and ($a,b$) are perpendicular $\Rightarrow (-2)\left(\frac{-4a}{b}\right)=-1\Rightarrow 8a=-b\Rightarrow b=-8a$

$4a^2+b^2=8,b=-8a\Rightarrow 4a^2+64a^2=8\Rightarrow 68a^2=8\Rightarrow a^2=2/17$.