 # A hyperbola having the transverse axis of length

Question:

A hyperbola having the transverse axis of length $\sqrt{2}$ has the same foci as that of the ellipse $3 x^{2}+4 y^{2}=12$, then this hyperbola does not pass through which of the following points ?

1. $\left(1,-\frac{1}{\sqrt{2}}\right)$

2. $\left(\sqrt{\frac{3}{2}}, \frac{1}{\sqrt{2}}\right)$

3. $\left(\frac{1}{\sqrt{2}}, 0\right)$

4. $\left(-\sqrt{\frac{3}{2}}, 1\right)$

Correct Option: , 2

Solution:

Ellipse : $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$

eccentricity $=\sqrt{1-\frac{3}{4}}=\frac{1}{2}$

$\therefore$ foci $=(\pm 1,0)$

for hyperbola, given $2 \mathrm{a}=\sqrt{2} \Rightarrow \mathrm{a}=\frac{1}{\sqrt{2}}$

$\therefore$ hyperbola will be

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

eccentricity $=\sqrt{1+2 b^{2}}$

$\therefore$ foci $=\left(\pm \sqrt{\frac{1+2 b^{2}}{2}}, 0\right)$

$\because$ Ellipse and hyperbola have same foci

$\Rightarrow \sqrt{\frac{1+2 b^{2}}{2}}=1$

$\Rightarrow \quad b^{2}=\frac{1}{2}$

$\therefore$ Equation of hyperbola : $\frac{x^{2}}{1 / 2}-\frac{y^{2}}{1 / 2}=1$

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

Clearly $\left(\sqrt{\frac{3}{2}}, \frac{1}{\sqrt{2}}\right)$ does not lie on it.