Tangent to Ellipse

The upper half of the ellipse is the graph of a concave function $f(x)$ . If $L(x)$ is the tangent at any point on the upper half, then $c=L(0)\geq f(0)=\sqrt{2}>1$ . By symmetry, for points on the lower half we have $c\leq -\sqrt{2}$ . Thus $c$ cannot be $\boxed{1}$ .

For being a tangent line should not touch the y axis in between $\sqrt2$ and $-\sqrt2$ otherwise it will be a secant. So c cannot be 1.

Given the Ellipse $\frac{x^2}{4}+\frac{y^2}{2}=1$ , the equation of tangent to ellipse is $\boxed{y=m x \pm \sqrt {a^2 m^2 +b^2}}$

Now, for $y=m x+c$ to be tangent to ellipse $\boxed{c=\pm \sqrt {a^2 m^2+b^2}}$

Substituting values of $a$ and $b$ , we get

$c=\sqrt {4 m^2+2}$

Since, $\sqrt{4 m^2+2} \geq \sqrt 2$ or $\sqrt {4 m^2+2}\leq -\sqrt 2$

$\rightarrow$ $c\geq \sqrt 2$ or $c\leq -\sqrt 2$

Hence $c$ cannot take $1$ as its value.