Subject: conics - finding axis of symmetry Name: Quinn Who is asking: Student Level: Secondary Question: I know the formula to find the axis of symmetry of a conic section (I'm not sure what shape - circle for the first one??) is (-D/2A,-E/2C) but I obviously don't get how to calculate it, because when I check the answer it's wrong, but I'm so close!! For the following equations my teacher suggested to "divide the x term coefficient, D, by the x squared term coefficient before...do the same for y." 2x2+2y2-8x+12y+16=0 what I did: new D=-4, E=6 (--4/2*2, -6/2*2) (1,-3) Real solution:(2,-3) x2+2y2-2x+8y-4=0 what I did: new D=-2, E=4 (--2/2*1, -4/2*2) (1,-1) real solution:(1,-2) I think this one is a parabola and the book says if A=0 y==E/2C; if C=0 x=-D/2A. Well ok but how do you find x when A=0 or y when C=0?? Help!! None of this is makes any sense! Hi Quinn, I would approach these problems another way, without using the standard form. For the first problem, 2x2 + 2y2 - 8x + 12y + 16 = 0 collect the terms involving x together and factor from these terms the coefficient of x2. Do the same for the terms involving y. 2x2 - 8x + 2y2 + 12y + 16 = 0 2(x2 - 4x) + 2(y2 + 6y) + 16 = 0 Now complete the square for both the x and y terms, adding the appropriate values to the right side to maintain the equality 2(x2 - 4x + 4) + 2(y2 + 6y + 9) + 16 = 8 + 18 The expression then becomes 2(x - 2)2 + 2(y + 3)2 = 26 - 16 (x - 2)2 + (y + 3)2 = 5 In this form the conic is recognizable as the circle with center (2,-3) and radius the square root of 5.    Try the same technique with the second problem. Cheers, Penny Go to Math Central