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. (a) the equation of the hyperbola is $\frac{x^2}{3}-\frac{y^2}{2}=1$
• B. (b) a focus of the hyperbola is (2, 0)
• C. (c) the eccentricity of the hyperbola is
• D. (d) the equation of the hyperbola is ${x^2}-3y^2=3$

Answer 1

Answer: (D)

(d) the equation of the hyperbola is ${x^2}-3y^2=3$

Answer 2

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 $(\pm \alpha e,0)\Rightarrow(\pm\sqrt{3},0)$
For hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,e^2_1=1+\frac{b^2}{a^2}$
where, $e^2_1=\frac{1}{e^2}=\frac{4}{3}\Rightarrow 1+\frac{b^2}{a^2}=\frac{4}{3}$
$\therefore$ $\frac{b^2}{a^2}=\frac{1}{3}$ ...........(i)
and hyperbola passes through $(\pm\sqrt{3},0)$
$\Rightarrow$ $\frac{3}{a^2}=1\Rightarrow \alpha^2=3$...........(ii)
From Eqs. (i) and (ii), $b^2-1$...................(iii)
$\therefore$ Equation of hyperbola is $\frac{x^2}{3}-{y^2}{1}=1$
Focus is $(\pm \alpha e_1,0)\Rightarrow \bigg(\pm\sqrt{3}.\frac{2}{\sqrt3,0}\bigg)\Rightarrow(\pm2,0)$
Hence, (b) and (d) are correct answers.