Question

Let the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ be reciprocal to that of the ellipse $x^{2}+4 y^{2}=4$. 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 $x^{2}-3 y^{2}=3$

Answer 1

Answer: (B), (D)

Ellipse is $\frac{x^{2}}{2^{2}}+\frac{y^{2}}{1^{2}}=1$
$1^{2}=2^{2}\left(1- e ^{2}\right)$
$\Rightarrow e =\frac{\sqrt{3}}{2}$
$\therefore $ eccentricity of the hyperbola is $\frac{2}{\sqrt{3}}$
$\Rightarrow b^{2}=a^{2}\left(\frac{4}{3}-1\right)$
$\Rightarrow 3 b^{2}=a^{2}$
Foci of the ellipse are $(\sqrt{3}, 0)$ and $(-\sqrt{3}, 0)$.
Hyperbola passes through $(\sqrt{3}, 0)$
$\frac{3}{a^{2}}=1 \Rightarrow a^{2}=3 \text { and } b^{2}=1$
$\therefore $ Equation of hyperbola is $x^{2}-3 y^{2}=3$
Focus of hyperbola is
$( ae , 0) \equiv\left(\sqrt{3} \times \frac{2}{\sqrt{3}}, 0\right) \equiv(2,0)$