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 $x^2+4y^2=4$. 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 $x^2-3y^2=3$

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 $\frac{2}{\sqrt{3}}$
$\Rightarrow b^2=a^2\left(\frac{4}{3}-1\right)$
$\Rightarrow 3b^2=a^2$
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 $x^2-3y^2=3$
Focus of hyperbola is
$(ae,0)\equiv \left(\sqrt{3}\times \frac{2}{\sqrt{3}},0\right)\equiv (2,0)$