Principal Axes of a Conic Section

It is known that curves of the form

Ax2+Bxy+Cy2 = k

(2)

are either circles, ellipses, or hyperbolas. If the curve is an ellipse or a hyperbola, then the points closest to and farthest from the origin lie on the principal axes of the conic.

Thus, the principal axes can be determined by finding the extrema of f( x,y) = x²+y² subject to the constraint (2). Moreover, the equivalent form Ñg = lÑf is often used in finding the principal axes.       

EXAMPLE 4    Find the principal axes of the conic

5x2+4xy+2y2 = 21

Solution: Since g( x,y) = 5x²+4xy+2y², we obtain

Ñg = á 10x+4y,4x+4y ñ         and       Ñf = á 2x,2y ñ

As a result, Ñg = lÑf leads to 10x+4y = l2x,   4x+4y = l2y, which is the same as

5x+2y = lx,    2x+2y = ly

We multiply the first by y and the second by x to obtain

5xy+2y2 = lxy = 2x2+2xy

Thus, 5xy+2y² = 2x²+2xy so that 2y²+3xy-2x² = 0. Factoring leads to

( y+2x) ( 2y-x) = 0

so that either y = -2x or y = x/2. Thus, y = -2x and  y = x/2  are the principal axes since the extrema must occur on these lines. Indeed, substitution of y = -2x into the constraint yields

5x2+4x( 2x) +2( 2x) 2 = 21 5x2+8x2+8x2 = 21 21x2 = 21

so that x = 1 or -1. Thus, when y = -2x, the critical points are (1,-2) and ( -1,2) . Likewise, y = x/2 yields

5x2+4x( x/2) +2( x/2) 2 = 21

which yields x = Ö{70}/5 or -Ö{70}/5 and the critical points

æ ç è 70 5 , 70 10 ö ÷ ø   and   æ ç è - 70 5 , 70 10 ö ÷ ø

These points lie on either y = -2x or y = x/2.

If the conic is a hyperbola, then there is only one principal axis. Correspondingly, there will be only one pair of critical points for the Lagrange multiplier problem.       

EXAMPLE 5    Find the principal axes of the conic

x2-3xy+y2 = 5

Solution: Since g( x,y) = x²-3xy+y², we obtain

Ñg = á 2x-3y,-3x+2y ñ         and       Ñf = á 2x,2y ñ

As a result, Ñg = lÑf leads to 2x-3y = 2lx and -3x+2y = 2ly, which in turn yields

2xy-3y2 = lxy,    -3x2+2xy = lxy

Thus, 2xy-3y² = -3x²+2xy, which yields -3y² = -3x² and consequently, y² = x².
       If y = -x, then the constraint becomes

x2-3x( -x) +x2 = 5        Þ        5x2 = 5

Thus, x = 1, -1 implying the critical points ( 1,-1) and ( -1,1) . If y = x, then the constraint becomes

x2-3x2+x2 = 5        Þ         -x2 = 5

This has no solution, so that y = x is not an axis of the conic. Since the conic has only one principal axis, it must be a hyperbola.