Conic Sections
Problems
Problem :
Find the coordinates of the foci of the ellipse
6x
2 + 
xy + 7y
2 - 36 = 0
.
This ellipse has an xy -term, so we'll have to rotate the axes to eliminate that term and find the standard form of the ellipse in the x'y' coordinate system. Then we'll find the foci and convert back to (x, y) for the answer.
The axes must be rotated through an angle
θ
such that
cot(2θ) = 
.
= - 
. Therefore,
θ = 
.
Next we must convert the
x
and
y
coordinates to
x'
and
y'
coordinates in
the new coordinate system which is a rotation of the coordinate axes by
θ =
radians. These conversions are as follows:
x = x'cos(θ) - y'sin(θ)
, and
y = x'sin(θ) + y'cos(θ)
. Substituting
θ =
, we find that
x = 
, and
y = 
. Then these values for
x
and
y
are substituted
in the equation
6x
2 +
xy + 7y
2 - 36 = 0
. After a lot of messy
algebra, the equation simplifies to
30x'
2 +22y'
2 = 144
. This equation in
standard form is
+ 
= 1
.
a > b
, so we know that
a 
2.5584
and
b 2.1909
. Therefore
c 1.3211
. The major axis is vertical (based on the form of the
equation in which the
y
2
term is the numerator of the fraction whose
denominator is
a
2
). Therefore the foci are located at
(0,±1.3211)
.
Keep in mind that these are
(x', y')
coordinates, and not yet
(x, y)
coordinates. The
x'
and
y'
axes are rotated
radians in a
counterclockwise direction from the
x
and
y
axes. To find the
x
and
y
coordinates of the foci, we must convert
x'
and
y'
back to
x
and
y
. We
use the same equations as before, and eventually find out that the foci are
located at
(x, y) (- 1.144,.6605)
and
(1.144, - .6605)
. The
approximations were a result of square roots taken. This is how to rotate the
axes to eliminate the
xy
-term of a conic to get in into standard form.
