Conic sections - Hyperbola

Question

If we are given a 2 degree curve equation of a hyperbola, is there a way to find the centre, foci, eccentricity and directrices of a hyperbola , just as we can obtain the equation using foci, eccentricity, and directrix? I searched for it, but only found an answer for ellipse.

Answer 1

It is nearly the same equations as you would use for an ellipse.

If you can get it into this form

$\frac {(x-h)^2}{a^2} - \frac{(y-v)^2}{b^2} = 1$

then

center: $(h,v)$

vertices are at $(h\pm a, v)$

$c = \sqrt{a^2 + b^2}$

The eccentricity is $e = \frac {c}{a}$

The foci are at $(h\pm c, v)$ the directrix is at $x= h\pm \frac {a^2}{c}$

if $-\frac {(x-h)^2}{a^2} + \frac{(y-v)^2}{b^2} = 1$

Then some of the equations above will swap $a$ for $b$ and the changes will be happening on a vertical, instead of a horizontal axis.

Comparison to an ellipse

$\frac {(x-h)^2}{a^2} + \frac{(y-v)^2}{b^2} = 1$

if $a>b$

center: $(h,v)$

vertices: $(h\pm a, v)$

$c = \sqrt{a^2 - b^2}$

eccentricity: $e = \frac {c}{a}$

foci: $(h\pm c, v)$

directrix: $x = h\pm \frac {a^2}{c}$