Problem 143

To find the value of $a$ for which the hyperbola $\frac{x^2}{a}-\frac{y^2}{2}=1$ and the ellipse $\frac{x^2}{4}+\frac{y^2}{a^2}=1$ share the same foci, we need to understand the formulas for the foci of both conic sections.

For a hyperbola of the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, the distance of each focus from the center is $\sqrt{a^2+b^2}$. In our case, $a^2 = a$ and $b^2 = 2$, so the distance is $\sqrt{a+2}$.

For an ellipse of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the distance of each focus from the center is $\sqrt{a^2-b^2}$ if $a > b$. Here, $a^2 = 4$ and $b^2 = a^2$, so the distance is $\sqrt{4-a^2}$.

Since the hyperbola and the ellipse share the same foci, their distances from the center to a focus are equal:
$$\sqrt{a+2} = \sqrt{4-a^2}$$
Squaring both sides to eliminate the square root gives:
$$a + 2 = 4 - a^2$$
Rearranging the equation, we get:
$$a^2 + a - 2 = 0$$
Factoring the quadratic equation, we find:
$$(a + 2)(a - 1) = 0$$
So, $a = -2$ or $a = 1$. However, since the condition $2 > a > 0$ must be met (as the values pertain to lengths in the context of conic sections and must be positive), we discard $a = -2$ and accept $a = 1$.

Therefore, the correct answer is $\boxed{A}$.