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Let the eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be reciprocal to that of the ellips $x^2+4 y^2=4$. If the hyperbola passes through a focus of the ellipse, then
$(A)$ the equation of the hyperbola is $\frac{x^2}{3}-\frac{y^2}{2}=1$
$(B)$ a focus of the hyperbola is $(2,0)$
$(C)$ the eccentricity of the hyperbola is $\sqrt{\frac{5}{3}}$
$(D)$ the equation of the hyperbola is $x^2-3 y^2=3$
$(B,C)$
$(A,D)$
$(B,D)$
$(C,D)$
Solution
For the ellipse
$\frac{x^2}{2^2}+\frac{y^2}{1^2}=1$
we have $1^2=2^2\left(1-e^2\right)$
or $e=\frac{\sqrt{3}}{2}$
Therefore, the eccentricity of the hyperbola is $2 \sqrt{3}$. So, for hyperbola $b^2=a^2\left(\frac{4}{3}-1\right)$ ItBrgt or $3 b^2=a^2$
One of the foci of the ellipse is $(\sqrt{3}, 9)$.
Therefore, $\frac{3}{a^2}=1$
or $a^2=3$ and $b^2=1$
Therefore, the equation of the hyperbola is $\frac{x^2}{3}-\frac{y^2}{1}=1$ ItBrgt the focus of the hyperbola is
$(a e, 0) \equiv\left(\sqrt{3} \times \frac{2}{\sqrt{3}}, 0\right) \equiv(2,0)$
