Subject: conics - finding axis of symmetry
Name: Quinn
Who is asking: Student
Level: Secondary

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 the same for y."

what I did:
new D=-4, E=6
(--4/2*2, -6/2*2)
Real solution:(2,-3)

what I did:
new D=-2, E=4
(--2/2*1, -4/2*2)
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.


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