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 $(\pm a e, 0) \Rightarrow(\pm \sqrt{3}, 0)$
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 $(\pm \sqrt{3}, 0)$
$\Rightarrow \frac{3}{a^{2}}=1 \Rightarrow a^{2}=3 ....$(ii)
From Eqs. (i) and (ii), $b^{2}=1 .....$(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.