Question

A hyperbola having the transverse axis of length $\sqrt{2}$ is confocal with $3x^2+4y^2=12$, then its equation is:

• A. $2x^2-2y^2=1$
• B. $2x^2+2y^2=1$
• C. $x^2+y^2=2$
• D. $x^2-y^2=2$

Answer 1

The correct option is A $2x^2-2y^2=1$

Given ellipse may be written as, $\frac{x^2}{4}+\frac{y^2}{3}=1$
$\Rightarrow a^2=4,b^2=3\therefore e=\sqrt{1-\frac{3}{4}}=\frac{1}{2}$
Thus foci of ellipse $(\pm 1,0)$
Hence foci of the hyperbola is $(\pm 1,0)$
And semi-major axis $a=\frac{1}{\sqrt{2}}$
$\Rightarrow \pm 1=\sqrt{a^2+b^2}\Rightarrow b^2=\frac{1}{2}$
or $b^2=1-a^2=1-\frac{1}{2}=\frac{1}{2}$
The required equation of hyperbola is,
$\frac{x^2}{1/2}-\frac{y^2}{1/2}=1\Rightarrow 2x^2-2y^2=1$