If the foci of frac{x^{2}}{16}+frac{y^{2}}{4} | vedclass

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(A) For the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$,we have $a^2=16$ and $b^2=4$. The eccentricity $e_1$ is given by $b^2 = a^2(1-e_1^2)$,so $4 =

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$\sqrt{3}$

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(A) For the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$,we have $a^2=16$ and $b^2=4$. The eccentricity $e_1$ is given by $b^2 = a^2(1-e_1^2)$,so $4 = 16(1-e_1^2)$,which gives $1-e_1^2 = \frac{1}{4}$,so $e_1^2 = \frac{3}{4}$,$e_1 = \frac{\sqrt{3}}{2}$. The foci are $(\pm a_1 e_1, 0) = (\pm 4 \cdot \frac{\sqrt{3}}{2}, 0) = (\pm 2\sqrt{3}, 0)$. For the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{9}=1$,we have $A^2=a^2$ and $B^2=9$. The eccentricity $e_2$ is given by $B^2 = A^2(e_2^2-1)$,so $9 = a^2(e_2^2-1) = a^2 e_2^2 - a^2$. The foci are $(\pm A e_2, 0) = (\pm a e_2, 0)$. Since the foci coincide,$a e_2 = 2\sqrt{3}$,so $a^2 e_2^2 = 12$. Substituting this into the hyperbola equation: $9 = 12 - a^2$. Thus,$a^2 = 3$,which implies $a = \sqrt{3}$.

If the foci of $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$ and $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{9}=1$ coincide,then the value of $a$ is

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