Exercises

Precalculus for Everybody

Chapter 3. The Conics

General 2nd Degree Equation. Rotations

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.

  1. \[ \left( 1, {-\sqrt{3}} \right),\,60^{\circ} \]
  2. \[ (-2, 6),\,45^{\circ} \]
  3. \[ \left( {-2\sqrt{3}}, 4 \right),\,30^{\circ} \]

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.

  1. \[ \left( {2-\sqrt{3}},\,{-1-2\sqrt{3}} \right),\, 60^{\circ} \]
  2. \[ \left( -3+{\frac{3\sqrt{2}}{2}},\, -3-{\frac{3\sqrt{2}}{2}} \right),\,45^{\circ} \]

In the exercises, from 6 to 7, find the transformed equation if the XY system is rotated at the indicated angle. Identify the conic.

  1. \[ 2xy=-1, \; \frac{\pi}{4} \text{ rad} \]
  2. \[ x^2+4\sqrt{3}xy-3y^2=30, \; \frac{ \pi}{6} \text{ rad} \]

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.

  1. \[ 2x^2+\sqrt{3}xy+y^2=5 \]
  2. \[ 9x^2+12xy+4y^2+2\sqrt{13}x-3\sqrt{13}y=0 \]
    \[ \begin{aligned} 9x^2+12xy+4y^2&+2\sqrt{13}x \\[1em] &-3\sqrt{13}y=0 \end{aligned} \]

  3. Let \(13x^2 -8xy + 7y^2 - 45 = 0\)

    1. By rotation of axes, verify that the graph of the equation is an ellipse.
    2. Find the vertices in both, X\('\)Y\('\) and XY, coordinate system.
    3. Find the foci in both, X\('\)Y\('\) and XY, coordinate system.
    4. Find the line containing the major axis in the XY system.
    5. Find the line containing the minor axis in the XY system.

  4. Let \(4x^2 -24xy + 11y^2 + 56x - 58y + 95 = 0\)

    Let:

    \[ \begin{aligned} 4x^2 -24xy + 11y^2 &+ 56x \\[1em] &- 58y + 95 = 0 \end{aligned} \]
    1. By a rotation of axes verify that the graph of the equation is a hyperbola.
    2. Find the center in both, the X\('\)Y\('\) and XY, coordinate system.
    3. Find the foci in both, the X\('\)Y\('\) and XY, coordinate system.
  1. \[ \left( -1, \, -\sqrt{3} \right) \]
  2. \[ \left( 2 \sqrt{2}, \, 4 \sqrt{2} \right) \]
  3. \[ \left( -1, \, 3 \sqrt{3} \right) \]
  4. \[ (4, \, -2) \]
  5. \[ \left( 3, \, -3 \sqrt{2} \right) \]
  6. \[ {y’}^2 – {x’}^2 = 1, \; \text{ hyperbola} \]
  7. \[ \frac{ {y’}^2 }{10} – \frac{ {x’}^2 }{6} = 1, \; \text{hyperbola} \]
  8. \[ \text{Ellipse}, \; \frac{ {x’}^2 }{2} + \frac{{y’}^2}{10} = 1 \]
  9. \[ \text{Parabola}, \; y’ = {x’}^2 \]
  10.  

    1. \(\cfrac{{x’}^2}{9} + \cfrac{{y’}^2}{3} = 1\)

    2. In X’Y’:   \(V_1 = (-3, \, 0)\),   \(V_2 = (3, \, 0)\).

      In XY:   \(V_1 = \left( -\frac{3}{\sqrt{5}}, \, -\frac{6}{\sqrt{5}} \right)\),   \(V_2 = \left( \frac{3}{\sqrt{5}}, \, \frac{6}{\sqrt{5}} \right)\)

    3. In X’Y’:   \(F_1 = \left( -\sqrt{6}, \, 0 \right)\),   \(F_2 = \left( \sqrt{6}, \, 0 \right)\).

      In XY:   \(F_1 = \left( -\sqrt{\frac{6}{5}}, \, -2\sqrt{ \frac{6}{5} } \right)\),   \(F_2 = \left( \sqrt{ \frac{6}{5} }, \, 2 \sqrt{ \frac{6}{5} } \right)\).

    4. \(2x – y = 0\)

    5. \(x + 2y = 0\)

  11.  

    1. \(\cfrac{(x’ – 1)^2}{4} + \cfrac{ (y’ – 2)^2 }{1} = 1\)

    2. In X’Y’:   \(C = (1, \, 2)\),   \(V_1 = (-1, \, 2)\),   \(V_2 = (3, \, 2)\).

      In XY:   \(C = \left( -\frac{2}{5}, \, \frac{11}{5} \right)\),   \(V_1 = (-2, \, 1)\),   \(V_2 = \left( \frac{6}{5}, \, \frac{17}{5} \right)\).

    3. In X’Y’:   \(F_1 = \left( 1 – \sqrt{5}, \, 2 \right)\),   \(F_2 = \left( 1 + \sqrt{5}, \, 2 \right)\)

      In XY:   \(F_1 = \left( – \frac{ 2 + 4 \sqrt{5} }{5}, \, \frac{11 – 3 \sqrt{5}}{5} \right)\),   \(F_2 = \left( -\frac{2 – 4 \sqrt{5}}{5}, \, \frac{11 + 3 \sqrt{5}}{5} \right)\)