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. $\pm 1=\pm \sqrt{a^2+b^2}$
$\Rightarrow b^2=1-a^2=1-\frac{1}{2}=\frac{1}{2}$
Hence, the equation of the required hyperbola is $\frac{x^2}{\frac{1}{2}}-\frac{y^2}{\frac{1}{2}}=1$