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
• D. the equation of the hyperbola is $x^2-3y^2=3$

Answer 1

Answer: (B), (D)

Here, equation of ellipse is $\frac{x^2}{4}+\frac{y^2}{1}=1$ $\Rightarrow e^2=1-\frac{b^2}{a^2}=1-\frac{1}{4}=\frac{3}{4}$
$\therefore e=\frac{\sqrt{3}}{2}$ and focus $(\pm ae,0)\Rightarrow (\pm \sqrt{3},0)$
For hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,e_1^2=1+\frac{b^2}{a^2}$
where, $e_1^2=\frac{1}{e^2}=\frac{4}{3}\Rightarrow 1+\frac{b^2}{a^2}=\frac{4}{3}$
$\therefore \frac{b^2}{a^2}=\frac{1}{3}....$(i)
and hyperbola passes through $(\pm \sqrt{3},0)$
$\Rightarrow \frac{3}{a^2}=1\Rightarrow a^2=\mathrm{3....}$(ii)
From Eqs. (i) and (ii), $b^2=\mathrm{1.....}$(iii)
$\therefore$ Equation of hyperbola is $\frac{x^2}{3}-\frac{y^2}{1}=1$
Focus is $\left(\pm a{e}_1,0\right)\Rightarrow \left(\pm \sqrt{3}\cdot \frac{2}{\sqrt{3}},0\right)\Rightarrow (\pm 2,0)$
Hence, (b) and (d) are correct answers.