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 equation of the hyperbola is $x^2–3y^2=3$
• D. The eccentricity of the hyperbola is $\sqrt{\frac{5}{3}}$

Answer 1

Answer: (B), (C)

The equation of the hyperbola is $x^2–3y^2=3$

The equation of the given ellipse is $\frac{x^2}{4}+\frac{y^2}{1}=1$
Whose eccentricity $e=\frac{\sqrt{3}}{2}Fociofellipseare(\sqrt{3},0$ and $(-\sqrt{3},0)$
Eccentricity of hyperbola $=\frac{2}{\sqrt{3}}$
Since given hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ passes through $(\sqrt{3},0),$
Hence $\frac{3}{a^2}-\frac{0}{b^2}=1$
$\Rightarrow a^2=3$
$b^2=a^2(e^2-1)$
Thus, the equation of the hyperbola is

$\frac{x^3}{3}-\frac{y^2}{1}=1$
$\Rightarrow x^2-3y^2=3$

Foci of the hyperbola are (\pm 2, 0\).