A student planed to investigate the graph of Ax 2 + Cy 2 + Dx +Ey -56 =0 A is ra

Question

A student planed to investigate the graph of Ax2 + Cy2 + Dx +Ey -56 =0

A is randomly selected from ( -1,2,3)  

C is randomly selected from ( -3,-2,0,6) 

D is randomly selected from ( -1,0,1,3)

 E is selected from (-3,-1,0,1)

a) Determine the number of distinct equations using all possible given values of A,C,D and E

 

b) determine the number of parabola equations that the graph produces 

 

c) Determine the number of hyperbola equations that the graph produces

Explanation / Answer

a) Multiply the # of possibilities together: 3*4*4*4 = 192 distinct equations. b) ax^2 + bxy + cy^2 + dx + ey + f This is the standard form for a bivariate (two variable) quadratic equation. D = b^2 - 4ac If D < 0, it is an ellipse. If D = 0, it is a parabola. If D > 0, it is a hyperbola. Here though, b = 0 (no b value). Possible A values are (-1, 2, 3) and possible C values are (-3, -2, 0, 6). The only way D could equal 0 is if A or C was also zero. This can happen if C = 0 and A could be either -1, 2 or 3. This can happen at all values of D and E as well. Multiplying the possibilities together (A=3, C=1, D=4, E=4) we get 48 parabola equations. c) D = -4ac with A values being (-1,2,3) and C being (-3,-2,0,6). D needs to be greater than 0 for a hyperbola. ONLY ONE of the A or C values can be positive, and neither can be zero. Possibilities include, in a,c format: (-1,6), (2,-3), (2,-2), (3,-3), and (3,-2). Multiplying the possibilities together, with a+c = 5, d=4, e=4, we get 80 hyperbola equations.

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