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 = 0

and 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} = 1

which is the equation of an ellipse, as shown in Figure 4.22.

4.22

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