Question 4.8.7: Rotation of a Conic Section Perform a rotation of axes to el...
Rotation of a Conic Section
Perform a rotation of axes to eliminate the xy-term in
5x² – 6xy + 5y² + 14√2x – 2√2y + 18 = 0and sketch the graph of the resulting equation in the x´y´-plane.
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The angle of rotation is given by
cot 2θ = \frac{a – c}{b} = \frac{5 – 5}{-6} = 0.
This implies that θ = π/4. So,
sin θ = \frac{1}{√2} and cos θ = \frac{1}{√2}.
By substituting
x = x´ cos θ – y´ sin θ = \frac{1}{√2}(x´ – y´)
and
y = x´ sin θ + y´ cos θ = \frac{1}{√2}(x´ + y´)
into the original equation and simplifying, you obtain
(x´)² + 4(y´)² + 6x´ – 8y´ + 9 = 0.Finally, by completing the square, you find the standard form of this equation to be
\frac{(x´ + 3)²}{2²} + \frac{(y´ – 1)²}{1²} = \frac{(x´ + 3)²}{4} + \frac{(y´ – 1)²}{1} = 1which is the equation of an ellipse, as shown in Figure 4.22.

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