Conic SectionsThe ProblemWe all know how to sketch the graph of an ellipse or hyperbola given an equationsuch as

x2 + 2y2 + 4x + 12y + 6 = 0by completing the square, nding the center of the ellipse, plotting the vertices andsketching the graph. If you forgot this process click here. The goal of this discussionis to be able to sketch a conic section given by ax2 + 2bxy + cy2 + dx + ey + f = 0The fact that there is an xy term here poses the challenge. We will look at theexample

Our goal is to graph this conic. At this point, we do not even know if it is a hyperbola,ellipse, or circle. The type will come out soon enough.

The SolutionThe key to solving this is to realize this as a matrix equation: xTAx + Bx + f = 0Where A is a 2 x 2 matrix, B is a 1 x 2 matrix and xT = (x,y). For our example wehave

Next we diagonalize the matrix A. Since the matrix is symmetric, it can be orthogonally diagonalized. Theeigenvalues are 7 and 2. We can check that the diagonalization is given by

A = PDPTor

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We can write

xTAx = xTPDPTx = (PTx)TD(PTx)and let

y = PTx or x = PyThat is

and

So that

We can put this all together to get

7x'2 + 2y'2 + 6x' +8y' - 2/7 = 0Completing the square gives

7(x' + 3/7)2 + 2(y' + 2)2 = 9This is the ellipse centered at (-3/7, -2) in the (x', y') coordinate system. with

a = 3 / b = 3 /

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The effect of the change of coordinates is a rotation of the graph by an angle of arctan(-2) = -1.1

The value of -2 was obtained by finding P21/P11. The graph is shown below.

Eigenvalues and Types of ConicsIn algebra, we learn how to identify conics. If the signs are opposite then we get ahyperbola and if the signs are alike, but the coecients are dierent, we get anellipse. If the signs are the same, we get a circle. If there is an absence of a squaredterm, then we have a parabola. We have a similar classication system for ourrotated conics.TheoremLet xTAx + Bx + f = 0be a conic and let s = l1 l2then

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If A is the zero matrix, we have a line. OtherwiseIf s < 0 the conic is a hyperbolaIf s > 0 then we get an ellipse (possibly degenerate) that is a circle ifthe eigenvalues are equal.If s = 0 then we get a parabola

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