Determining the Type and Position of a Second-Order Curve from Its Equation

$\frac{x^{2}}{25}-\frac{y^{2}}{144} = 1$;

$\frac{x^{2}}{225}-\frac{y^{2}}{64} = -1$.

$\frac{x^{2}}{4}+\frac{y^{2}}{20} = 1$;

$\frac{x^{2}}{4}-2 y^{2}+8 = 0$;

$y = \frac{3}{4} x^{2}$.

$x^{2}+2 y^{2}+4 x-4 y = 0$

$4 x^{2}-y^{2}-8 x-6 y-4 = 0$;

$4 x^{2}+4 x+2 y-1 = 0$

$6 x^{2}+8 y^{2}+3 x-4 y+1 = 0$;

$2 x^{2}-3 y^{2}-6 x+9 y-2 = 0$;

$2 x^{2}+6 x+3 y+6 = 0$;

$4 y^{2}+4 x-2 y+1 = 0$;

$3 x^{2}-2 y^{2}+6 x+4 y+1 = 0$.

$5 x^{2}+8 x y+5 y^{2}-18 x-18 y+11 = 0$;

$2 x y-4 x+2 y+11 = 0$;

$4 x^{2}+4 x y+y^{2}-10 x-5 y+6 = 0$;

$x^{2}-2 x y+y^{2}-3 x+2 y-11 = 0$.

$\frac{(x+y-1)^{2}}{2}+\frac{(x-y+3)^{2}}{8}+1 = 0$;

$(x+2 y+1)^{2}+4 x-2 y-6 = 0$;

$(x+2 y+1)^{2}+x-3 y-6 = 0$;

$\frac{(x-3 y+2)^{2}}{10}-\frac{(3 x+y+1)^{2}}{20}+20 x = 0$.

$5 x^{2}+4 x y+8 y^{2}-32 x-56 y+80 = 0$;

$5 x^{2}+8 x y+5 y^{2}-18 x-18 y+9 = 0$;

$5 x^{2}+6 x y+5 y^{2}-16 x-16 y-16 = 0$;

$12 x y+5 y^{2}-12 x-22 y-19 = 0$;

$7 x^{2}+16 x y-23 y^{2}-14 x-16 y-218 = 0$;

$7 x^{2}-24 x y-38 x+24 y+175 = 0$;

$9 x^{2}+24 x y+16 y^{2}-40 x+30 y = 0$;

$x^{2}+2 x y+y^{2}-8 x+4 = 0$

$4 x^{2}-4 x y+y^{2}-2 x-14 y+7 = 0$.

Reduce the given equation to canonical form.

Find the relation between the old coordinates $x$ and $y$ and the new (canonical) coordinates $x^{\prime }$ and $y^{\prime }$.

Find the coordinates of the center, vertices, and foci of this curve relative to the original coordinate system.

Set up the equations of the directrices, asymptotes, and tangents at the vertices of this curve in the original coordinate system.