# How to find centre,vertics,foci,focal radii,letus rectum… when exists of a general quadratic equation in x and y

Is there a generalized way( a particular conic section of any shape,for instance an ellipse without determining its major/minor axis) to find the centre,vertices,foci,focal radii,letus rectum,eccentricity,etc when exists of any parabola,ellipse,hyperbola?(dealing separately)

$ax^2+by^2+cx+dy+exy+c=0$ ($a,b,c$ not all zero)

• Well, I think you first must know whether the conic is a circle, parabola, hyperbola...and yes: there are ways to know this, in particular by means of linear algebra, matrices (signature, dterminant and stuff), etc. – DonAntonio Apr 16 '14 at 11:58

One can use a list of equations to determine the property you require. Note, however, that in many cases it is easier to use derived value(s) as opposed to using equations that rely solely on the coefficients of the equation in general quadratic form. Below are examples of equations one can use.

Properties of an ellipse from equation for conic sections in general quadratic form

Given the equation for conic sections in general quadratic form:

$a x^2 + b x y + c y^2 + c x + e y + f = 0$

The equation represents an ellipse if:

$b^2 - 4 a c < 0$

or similarly,

$4 a c - b^2 > 0$

The coefficient normalizing factor is given by:

$q = 64 {{f (4 a c - b^2) - a e^2 + b d e - c d^2} \over {(4ac - b^2)^2}}$

The distance between center and focal point (either of the two) is given by:

$s = {1 \over 4} \sqrt { |q| \sqrt { b^2 + (a - c)^2 }}$

The semi-major axis length is given by:

$r_\max = {1 \over 8} \sqrt { 2 |q| {\sqrt{b^2 + (a - c)^2} - 2 q (a + c) }}$

The semi-minor axis length is given by:

$r_\min = \sqrt {{r_\max}^2 - s^2}$

The latus rectum is given by:

$l = 2 {{ {r_\min}^2 } \over {r_\max}}$

The eccentricity is given by:

$g = {{s} \over {r_\max}}$

The distance between center and closest directix point (either of the two) is given by:

$h = {{{r_\max}^2} \over {s}}$

The center of the ellipse is given by:

$x_\Delta = { b e - 2 c d \over 4 a c - b^2}$

$y_\Delta = { b d - 2 a e \over 4 a c - b^2}$

The top-most point on the ellipse is given by:

$y_T = y_\Delta + {\sqrt {(2 b d - 4 a e)^2 + 4(4 a c - b^2)(d^2 - 4 a f)} \over {2(4 a c - b^2)}}$

$x_T = {{-b y_T - d} \over {2 a}}$

The bottom-most point on the ellipse is given by:

$y_B = y_\Delta - {\sqrt {(2 b d - 4 a e)^2 + 4(4 a c - b^2)(d^2 - 4 a f)} \over {2(4 a c - b^2)}}$

$x_B = {{-b y_B - d} \over {2 a}}$

The left-most point on the ellipse is given by:

$x_L = x_\Delta - {\sqrt {(2 b e - 4 c d)^2 + 4(4 a c - b^2)(e^2 - 4 c f)} \over {2(4 a c - b^2)}}$

$y_L = {{-b x_L - e} \over {2 c}}$

The right-most point on the ellipse is given by:

$x_R = x_\Delta + {\sqrt {(2 b e - 4 c d)^2 + 4(4 a c - b^2)(e^2 - 4 c f)} \over {2(4 a c - b^2)}}$

$y_R = {{-b x_R - e} \over {2 c}}$

The angle between x-axis and major axis is given by:

if $(q a - q c = 0)$ and $(q b = 0)$ then $\theta = 0$
if $(q a - q c = 0)$ and $(q b > 0)$ then $\theta = {1 \over 4} \pi$
if $(q a - q c = 0)$ and $(q b < 0)$ then $\theta = {3 \over 4} \pi$
if $(q a - q c > 0)$ and $(q b >= 0)$ then $\theta = {1 \over 2} {atan ({b \over {a - c}})}$
if $(q a - q c > 0)$ and $(q b < 0)$ then $\theta = {1 \over 2} {atan ({b \over {a - c}})} + {\pi}$
if $(q a - q c < 0)$ then $\theta = {1 \over 2} {atan ({b \over {a - c}})} + {1 \over 2}{\pi}$

The focal points are given by:

$F_{1,x} = x_\Delta - s \ cos (\theta)$

$F_{1,y} = y_\Delta - s \ sin (\theta))$

$F_{2,x} = x_\Delta + s \ cos (\theta)$

$F_{2,y} = y_\Delta + s \ sin (\theta))$

• User R Nogueira claimed in a now-deleted answer that in the expression for $q$, the sign of the term $cd^2$ should be flipped. Symmetry considerations seem to back that up. – epimorphic Feb 8 '17 at 4:33