Question

4.(25 Pt) Let 2 - 2xy + y + 42.r - 4 be a quadratic form. a) Determine the kind of this quadratic form, b) Write in the standart form of this quadratic form c) Skecht the graph of this quadratic form

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Answer 1

1 -1 212 1 (262 o 4 Here the given general equation of and - 2xy + y +4888 = 4 degree, Compare this equation to the general equation of and degree, axs shay by + 2g x + 2 fy+C =0. So, aal, h=-1, bal, g = 252; f20, @2.4. Now, Da ab h = 1.1-2 1-120 0 = -4+4+27(0-20) I fe 2-8 (+0) So, here, the giren general equation of and degree represents a hyperbela, since, 220 & 4to. Now, The giren guertion may be written as, ar zxy & y +482x = 4 7 (-y) 4-419% » (x-y) 4-42% + 8 + 2xx-24 here, & being any Constant. (x-3 + x) = (22-482) & - 204 + 4 + 4. here, & is choosen that two straight line one penarbola are axis of the perrombola and another tingent at the vertex of the perpendiculow ie , x-gth 20 (21-42) &- 237 + 8 + 4 = 0 These two straight lines are perpendicular so, 2A-452 X 1 =-1. 21 21-4022-2d 41 2412 > 212 Most a

The giren equation The eguertion becomes, (x-y + 2) = (252-4V2)* - 2824 +8+4 *(x-+vB) = -270X-219 4 + 12. (-19,402) - 212 (x + y + 2)2 (x+y=3V2) - 282 (+9 .382) (1122 -2re (x+y = 352 2). a-2 +2 12 22 Y Y = - 2x xty-352 12 x-y +2 2 3 These are the axes of co-orolinates transformation so sportion of porrabola is, Y2 - 2x. represents a parabola whose axis, x-y 12 20 the equation of the fourgent at the vertex is x+y-35220 and the Jength of latus rectum is 2. The graph of ② is Taxty-352 2x Xox y tue 4 #ry+12=0 2 -y-302 123 12345 x+y=3V2=0