To lead the equation of curves of the second order to a canonical form. To find coordinates of focuses, to make the drawing. a) x^2-4y^2=16; b) x^2+y^2-x-y-0.5=0; v) 2x^2-3y^2=12; gy+x^2+4=0;

34
A) x^2 - 4y^2 = 16 x^2/16 - y^2/4 = 1 Hyperbole with a half shafts = 4; b = 2 c = √ (a^2+b^2) = √ (16+4) = √20 Focuses of F1 (-√20; 0); F2 (√20; 0) b) x^2 + y^2 - x - y - 0.5 = 0 x^2 - 2*x*0.5 + (0.5) ^2 + y^2 - 2*y*0.5 + (0.5) ^2 - 0.25 - 0.25 - 0.5 = 0 (x - 0.5) ^2 + (y - 0.5) ^2 = 1 Circle with the center (0.5; 0.5) and with a radius of 1. Focus, it is the center, F(0.5; 0.5) v) 2x^2 - 3y^2 = 12 x^2/6 - y^2/4 = 1 Hyperbole with a half shafts = √6; b = 2 c = √ (a^2+b^2) = √ (6 + 4) = √10 Focuses of F1 (-√10; 0); F2 (√10; 0) g) y + x^2 + 4 = 0 y + 4 =-x^2 the Parabola with top (0;-4) and parameter p =-0.5 F(0 Focus;-4.5)
169
Answer add