Graph the following equations.
\(x^2+2xy+y^2 -x\sqrt {2}+y\sqrt {2} -6= 0\)
\(x^2+2xy+y^2 -x\sqrt {2}+y\sqrt {2} -6= 0\) becomes \((x')^2 = -(y'-3)\) after rotating counter-clockwise through \(\theta = \frac {\pi }{4}\) .
\(7x^2-4xy\sqrt {3}+3y^2-2x-2y\sqrt {3}-5= 0\)
\(7x^2-4xy\sqrt {3}+3y^2-2x-2y\sqrt {3}-5= 0\) becomes \(\frac {(x'-2)^2}{9}+(y')^2 = 1\) after rotating counter-clockwise through \(\theta = \frac {\pi }{3}\)
\(5x^2+6xy+5y^2 - 4\sqrt {2}x+4\sqrt {2}y = 0\)
\(5x^2+6xy+5y^2 - 4\sqrt {2}x+4\sqrt {2}y = 0\) becomes \((x')^2+\frac {(y'+2)^2}{4} = 1\) after rotating counter-clockwise through \(\theta = \frac {\pi }{4}\) .
\(x^2+ 2\sqrt {3}xy+3y^2+ 2\sqrt {3}x-2y-16 = 0\)
\(x^2+ 2\sqrt {3}xy+3y^2+ 2\sqrt {3}x-2y-16 = 0\) becomes\((x')^2 = y'+4\) after rotating counter-clockwise through \(\theta = \frac {\pi }{3}\)
\(13x^2-34xy\sqrt {3}+47y^2 - 64=0\)
\(13x^2-34xy\sqrt {3}+47y^2 - 64=0\) becomes \((y')^2 - \frac {(x')^2}{16} =1 \) after rotating counter-clockwise through \(\theta = \frac {\pi }{6}\) .
\(x^2-2\sqrt {3} xy-y^2+8=0\)
\(x^2-2\sqrt {3} xy-y^2+8=0\) becomes \(\frac {(x')^2}{4} - \frac {(y')^2}{4} = 1\) after rotating counter-clockwise through \(\theta = \frac {\pi }{3}\)
\(x^2-4xy+4y^2-2x\sqrt {5}-y\sqrt {5}=0\)
\(x^2-4xy+4y^2-2x\sqrt {5}-y\sqrt {5}=0\) becomes \((y')^2=x\) after rotating counter-clockwise through \(\theta = \arctan \left (\frac {1}{2}\right )\) .
\(8x^2+12xy+17y^2 - 20 = 0\)
\(8x^2+12xy+17y^2 - 20 = 0\) becomes \((x')^2 + \frac {(y')^2}{4} = 1\) after rotating counter-clockwise through \(\theta = \arctan (2)\) .
Graph the following equations.
\(r = \frac {2}{1-\cos (\theta )}\)
\(r = \frac {2}{1-\cos (\theta )}\) is a parabola
directrix: \(x = -2\)
vertex: \((-1,0)\)
focus: \((0,0)\)
focal diameter: \(4\)
\(r = \frac {3}{2 + \sin (\theta )}\)
\(r = \frac {3}{2 + \sin (\theta )} = \frac {\frac {3}{2}}{1 + \frac {1}{2} \sin (\theta )}\) is an ellipse
directrix: \(y = 3\)
vertices: \((0,1)\) , \((0,-3)\)
center: \((0,-2)\)
foci: \((0,0)\) , \((0,-2)\)
minor axis length: \(2\sqrt {3}\)
\(r = \frac {3}{2-\cos (\theta )}\)
\(r = \frac {3}{2 - \cos (\theta )} = \frac {\frac {3}{2}}{1 - \frac {1}{2} \cos (\theta )}\) is an ellipse
directrix: \(x = -3\)
vertices: \((-1,0)\) , \((3,0)\)
center: \((1,0)\)
foci: \((0,0)\) , \((2,0)\)
minor axis length: \(2\sqrt {3}\)
\(r = \frac {2}{1 + \sin (\theta )}\)
\(r = \frac {2}{1+\sin (\theta )}\) is a parabola
directrix: \(y=2\)
vertex: \((0,1)\)
focus: \((0,0)\)
focal diameter: \(4\)
\(r = \frac {4}{1+3\cos (\theta )}\)
\(r = \frac {4}{1+3\cos (\theta )}\) is a hyperbola
directrix: \(x = \frac {4}{3}\)
vertices: \((1,0)\) , \((2,0)\)
center: \(\left (\frac {3}{2}, 0\right )\)
foci: \((0,0)\) , \((3,0)\)
conjugate axis length: \(2\sqrt {2}\)
\(r = \frac {2}{1-2\sin (\theta )}\)
\(r = \frac {2}{1-2\sin (\theta )}\) is a hyperbola
directrix: \(y = -1\)
vertices: \(\left (0,-\frac {2}{3}\right )\) , \((0,-2)\)
center:\(\left (0, -\frac {4}{3} \right )\)
foci: \((0,0)\) , \(\left (0, -\frac {8}{3}\right )\)
conjugate axis length: \(\frac {2\sqrt {3}}{3}\)
\(r = \frac {2}{1 + \sin (\theta - \frac {\pi }{3})}\)
\(r = \frac {2}{1 + \sin (\theta - \frac {\pi }{3})}\) is the parabola
\(r = \frac {2}{1 + \sin (\theta )}\) rotated through
\(\phi = \frac {\pi }{3}\) .
\(r = \frac {6}{3 - \cos \left (\theta + \frac {\pi }{4}\right )}\)
\(r = \frac {6}{3 - \cos \left (\theta + \frac {\pi }{4}\right )}\) is the ellipse
\(r = \frac {6}{3 - \cos \left (\theta \right )} = \frac {2}{1 - \frac {1}{3} \cos \left (\theta \right )}\) rotated through
\(\phi = -\frac {\pi }{4}\)
The matrix
\(A(\theta ) = \left [ \begin{array}{rr} \cos (\theta ) & -\sin (\theta ) \\ \sin (\theta ) & \cos (\theta ) \\ \end{array} \right ]\) is called a
rotation matrix .
We’ve seen this matrix most recently used in the proof of Theorem rotatecoordinatesthm .
Discuss with your classmates how to use \(A(\theta )\) to rotate points in the plane.
Using the even / odd identities for cosine and sine, show \(A(\theta )^{-1} = A(-\theta )\) . Interpret this geometrically.