Discussion Forum

Let the eccentricity of the hyperbola  $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}$ =1 be reciprocal to that of the ellipse x² + 4y² = 4. If the hyperbola passes through a focus of the ellipse, then

Answer:

Explanation:

 Here , equation of ellipse

$\frac{x^{2}}{4}+\frac{y^{2}}{1}=1$

 $\Rightarrow$   $e^{2}=1-\frac{b^{2}}{a^{2}}=1-\frac{1}{4}=\frac{3}{4}$

 $\therefore$   $e= \frac{\sqrt{3}}{2}$ and focus $(\pm ae,0)$

  $= ( \pm \sqrt{3},0)$

 For hyperbola    $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$

 $e_{1}^{2}=1+\frac{b^{2}}{a^{2}}$  where, $e_{1}^{2}=\frac{1}{e^{2}}=\frac{4}{3}$

$\Rightarrow$  $1+ \frac{b^{2}}{a^{2}}=\frac{4}{3} \Rightarrow \frac{b^{2}}{a^{2}}=\frac{1}{3}$....(i)

and hyperbola passes through $(\pm \sqrt{3},0)$

Now, $\frac{3}{a^{2}}=1 \Rightarrow  a^{2}=3$  ....(ii)

 from eqs. (i) and (ii) , we get

 $b^{2}=1$  ......(iiii)

 $\therefore$ Equation of hyperbola is 

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

 Focus is $(\pm ae,0)$

Now, $\left(\pm \sqrt{3}.\frac{2}{\sqrt{3}},0\right) \Rightarrow  (\pm2,0)$

 hence , both options (b) and (d)  are correct