In the exercises, from 1 to 3, the coordinates of a point are given in the XY system. The axes are rotated at the indicated angle. Find the coordinates of the point in the X\('\)Y\('\) system.
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\[ \left( 1, {-\sqrt{3}} \right),\,60^{\circ} \]
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\[ (-2, 6),\,45^{\circ} \]
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\[ \left( {-2\sqrt{3}}, 4 \right),\,30^{\circ} \]
In the exercises, from 4 to 5, the coordinates of a point are given in the X\('\)Y\('\) system, obtained by rotating the XY system at the indicated angle. Find the coordinates of the point in the XY system.
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\[ \left( {2-\sqrt{3}},\,{-1-2\sqrt{3}} \right),\, 60^{\circ} \]
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\[ \left( -3+{\frac{3\sqrt{2}}{2}},\, -3-{\frac{3\sqrt{2}}{2}} \right),\,45^{\circ} \]
In the exercises, from 6 to 7, find the transformed equation if the XY system is rotated at the indicated angle. Identify the conic.
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\[ 2xy=-1, \; \frac{\pi}{4} \text{ rad} \]
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\[ x^2+4\sqrt{3}xy-3y^2=30, \; \frac{ \pi}{6} \text{ rad} \]
In the exercises, from 8 to 9, use the discriminant to identify the conic. Use rotation of axes to eliminate the \(\boldsymbol{xy}\) term, find the transformed equation and plot it.
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\[ 2x^2+\sqrt{3}xy+y^2=5 \]
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\[ 9x^2+12xy+4y^2+2\sqrt{13}x-3\sqrt{13}y=0 \]\[ \begin{aligned} 9x^2+12xy+4y^2&+2\sqrt{13}x \\[1em] &-3\sqrt{13}y=0 \end{aligned} \]
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Let \(13x^2 -8xy + 7y^2 - 45 = 0\)
- By rotation of axes, verify that the graph of the equation is an ellipse.
- Find the vertices in both, X\('\)Y\('\) and XY, coordinate system.
- Find the foci in both, X\('\)Y\('\) and XY, coordinate system.
- Find the line containing the major axis in the XY system.
- Find the line containing the minor axis in the XY system.
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Let \(4x^2 -24xy + 11y^2 + 56x - 58y + 95 = 0\)
Let:
\[ \begin{aligned} 4x^2 -24xy + 11y^2 &+ 56x \\[1em] &- 58y + 95 = 0 \end{aligned} \]- By a rotation of axes verify that the graph of the equation is a hyperbola.
- Find the center in both, the X\('\)Y\('\) and XY, coordinate system.
- Find the foci in both, the X\('\)Y\('\) and XY, coordinate system.
Answers
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\[ \left( -1, \, -\sqrt{3} \right) \]
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\[ \left( 2 \sqrt{2}, \, 4 \sqrt{2} \right) \]
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\[ \left( -1, \, 3 \sqrt{3} \right) \]
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\[ (4, \, -2) \]
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\[ \left( 3, \, -3 \sqrt{2} \right) \]
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\[ {y’}^2 – {x’}^2 = 1, \; \text{ hyperbola} \]
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\[ \frac{ {y’}^2 }{10} – \frac{ {x’}^2 }{6} = 1, \; \text{hyperbola} \]
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\[ \text{Ellipse}, \; \frac{ {x’}^2 }{2} + \frac{{y’}^2}{10} = 1 \]
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\[ \text{Parabola}, \; y’ = {x’}^2 \]
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\(\cfrac{{x’}^2}{9} + \cfrac{{y’}^2}{3} = 1\)
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In X’Y’: \(V_1 = (-3, \, 0)\), \(V_2 = (3, \, 0)\).
In XY: \(V_1 = \left( -\frac{3}{\sqrt{5}}, \, -\frac{6}{\sqrt{5}} \right)\), \(V_2 = \left( \frac{3}{\sqrt{5}}, \, \frac{6}{\sqrt{5}} \right)\)
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In X’Y’: \(F_1 = \left( -\sqrt{6}, \, 0 \right)\), \(F_2 = \left( \sqrt{6}, \, 0 \right)\).
In XY: \(F_1 = \left( -\sqrt{\frac{6}{5}}, \, -2\sqrt{ \frac{6}{5} } \right)\), \(F_2 = \left( \sqrt{ \frac{6}{5} }, \, 2 \sqrt{ \frac{6}{5} } \right)\).
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\(2x – y = 0\)
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\(x + 2y = 0\)
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\(\cfrac{(x’ – 1)^2}{4} + \cfrac{ (y’ – 2)^2 }{1} = 1\)
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In X’Y’: \(C = (1, \, 2)\), \(V_1 = (-1, \, 2)\), \(V_2 = (3, \, 2)\).
In XY: \(C = \left( -\frac{2}{5}, \, \frac{11}{5} \right)\), \(V_1 = (-2, \, 1)\), \(V_2 = \left( \frac{6}{5}, \, \frac{17}{5} \right)\).
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In X’Y’: \(F_1 = \left( 1 – \sqrt{5}, \, 2 \right)\), \(F_2 = \left( 1 + \sqrt{5}, \, 2 \right)\)
In XY: \(F_1 = \left( – \frac{ 2 + 4 \sqrt{5} }{5}, \, \frac{11 – 3 \sqrt{5}}{5} \right)\), \(F_2 = \left( -\frac{2 – 4 \sqrt{5}}{5}, \, \frac{11 + 3 \sqrt{5}}{5} \right)\)
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