Question

Reduce the conic x
2 −6xy+y
2 −4 = 0 to standard form. Hence the given conic.

Answer 1

Answer on Question #66678 - Math - Linear Algebra

Question

Reduce the conic $x^{2} - 6xy + y^{2} - 4 = 0$ to standard form. Hence the given conic.

Solution

$$x^{2} - 6xy + y^{2} - 4 = 0$$ $$x = x' \cos \alpha - y' \sin \alpha$$ $$y = x' \sin \alpha + y' \cos \alpha$$ $$\tan 2\alpha = \frac{-6}{1 - 1} = \infty, \qquad 2\alpha = \frac{\pi}{2}, \qquad \alpha = \frac{\pi}{4}$$ $$x = \frac{\sqrt{2}}{2}(x' - y')$$ $$y = \frac{\sqrt{2}}{2}(x' + y')$$ $$\frac{1}{2}(x' - y')^{2} - 6 \times \frac{1}{2}(x' - y')(x' + y') + \frac{1}{2}(x' + y')^{2} - 4 = 0$$ $$x'^{2} + y'^{2} - 2x'y' - 6x'^{2} + 6y'^{2} + x'^{2} + y'^{2} + 2x'y' - 8 = 0$$ $$-4x'^{2} + 8y'^{2} = 8$$ $$\frac{y'^{2}}{1} - \frac{x'^{2}}{2} = 1$$

Answer: The curve is a hyperbole.

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