Question

Find Directrix, foci and eccentricity of the conics:

$x^2+2x-y^2+5=0$

• A. Directrices $y=\pm \sqrt{2}$
• B. foci $(-1,\pm 2\sqrt{2})$
• C. $e=\sqrt{2}$
• D. $e=2$

Answer 1

Answer: (A), (B), (C)

The correct options are
A Directrices $y=\pm \sqrt{2}$
B foci $(-1,\pm 2\sqrt{2})$
C $e=\sqrt{2}$

$x^2+2x-y^2+5=0$
$x^2+2x+1-y^2=-4$
$(x+1{)}^2-y^2=-4$
$\frac{y^2}{4}-\frac{(x+1{)}^2}{4}=1$
Clearly this equation of rectangular hyperbola with $y-$axis as major axis
eccentricity $e=\sqrt{2}$ directrices $:y=\frac{a}{e}=\pm \sqrt{2}$ foci $(-1,\pm ae)=(-1,\pm 2\sqrt{2})$