Problem : Find the coordinates of the foci of the ellipse 6x2 + xy + 7y2 - 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 6x2 + xy + 7y2 - 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 y2 term is the numerator of the fraction whose denominator is a2). 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.
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