How do I rotate the axes of and then graph $7 x^{2} - 6 \sqrt{3} x y + 13 y^{2} - 16 = 0$ $7x^2 - 6sqrt3xy + 13y^2 - 16 = 0$ ?>

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How do I rotate the axes of and then graph $7 x^{2} - 6 \sqrt{3} x y + 13 y^{2} - 16 = 0$ $7x^2 - 6sqrt3xy + 13y^2 - 16 = 0$ ?>

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How do you graph the conic $x^{2} - 4 x y + y^{2} + 1 = 0$ $x^2-4xy+y^2+1=0$ by first rotations the axis and eliminating the $x y$ $xy$ term?

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Answer 1 · 2016-05-01T17:26:25

The angle of rotation may be given as $θ = \frac{π}{6}$ $theta=pi/6$
The conic section is an ellipse and in standard form, may be given as $\frac{x^{2}}{4} + \frac{y^{2}}{1} = 1$ $x^2/4+y^2/1=1$

Explanation:

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Finally, we may graph this rotation in the xy plane.

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How do I rotate the axes of and then graph $7 x^{2} - 6 \sqrt{3} x y + 13 y^{2} - 16 = 0$ $7x^2 - 6sqrt3xy + 13y^2 - 16 = 0$ ?>

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Explanation:

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How do I rotate the axes of and then graph 7x2−6√3xy+13y2−16=07x^2 - 6sqrt3xy + 13y^2 - 16 = 0?>

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Explanation:

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How do I rotate the axes of and then graph $7 x^{2} - 6 \sqrt{3} x y + 13 y^{2} - 16 = 0$ $7x^2 - 6sqrt3xy + 13y^2 - 16 = 0$ ?>