Question

A hyperbola having the transverse axis of length $\sqrt{2}$ units has the same focii as that of ellipse $3x^{2}+4y^{2}=12$ , then its equation is

• A. $2x^{2}-2y^{2}=1$
• B. $2x^{2}-2y^{2}=3$
• C. $x^{2}-y^{2}=-2$
• D. $x^{2}-y^{2}=2$

Answer 1

Answer: (A)

For the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1,$ length of semi transverse axis $a=\frac{1}{\sqrt{2}},$
For the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1,$ the foci is $\left(\pm 1,0\right)$
i.e. $\pm1=\pm\sqrt{a^{2} + b^{2}}$
$\Rightarrow b^{2}=1-a^{2}=1-\frac{1}{2}=\frac{1}{2} \, $
Hence, the equation of the required hyperbola is $\frac{x^{2}}{\frac{1}{2}}-\frac{y^{2}}{\frac{1}{2}}=1$