Question

Find the eccentricity of the conic represented by $x^2-y^2-4x+4y+16=0$

• A. $\sqrt{2}$
• B. $\sqrt{3}$
• C. $-\sqrt{2}$
• D. $-\sqrt{3}$

Answer 1

The correct option is A $\sqrt{2}$

Given conic can be written as,

$x^2-4x+4-(y^2-4y+4)+16=0$

$(x-2{)}^2-(y-2{)}^2=-16$
$\frac{(x-2{)}^2}{16}-\frac{(y-2{)}^2}{16}=-1$
Which is a rectangular hyperbola so its eccentricity is $\sqrt{2}$