Question

The equation of the hyperbola whose eccentricity is $\sqrt{2}$ and the distance between the foci is $16$, taking tansverse and conjugate axes of the hyperbola as $x$ and $y$ axes respectively, is

• A. $x^2-y^2=0$
• B. $x^2-y^2=32$
• C. $x^2-y^2=2$
• D. None of these

Answer 1

Answer: (B)

$x^2-y^2=32$

Let the equation of the hyperbola be
$\frac{x^2}{a^2}=\frac{y^2}{b^2}=1$
The coordinates of the foci are
$(ae,0);(-ae,0)$
Therefore, $2ae=16\Rightarrow 2a\sqrt{2}=16\Rightarrow a=4\sqrt{2}$
Also, $b^2=a^2(e^2-1)=32(2-1)=32$
Thus, $a^2=32;b^2=32$
Hence, the required equation is
$\frac{x^2}{32}=\frac{y^2}{32}=1$
$\Rightarrow$ $x^2-y^2=32$