Chapter A5.3, Problem 30E

To determine

To find: a sine and cosine angle and say whether the equation is an ellipse, hyperbola or parabola.

Answer to Problem 30E

The new equation is a hyperbola.

The angle is $\alpha \approx {5.65}^{\circ }$ .

Sine and cosine angle is,

  $\begin{array}{l}\sin \alpha \approx 0.10\\ \cos \alpha \approx 0.995\end{array}$

Explanation of Solution

Given:

  $2x^2+xy-3y^2+3x-7=0$

Calculation:

Comparing with the standard equation for a quadratic curve,

  $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$

Thus,

  $A=2,B=1,C=-3$

For the cross-product term to vanish, the coordinate axes must be rotated by an angle $\alpha$ , such that ,

  $\begin{array}{l}\tan 2\alpha =\frac{B}{A-C}\\ =\frac{1}{2-(-3)}\\ =\frac{1}{5}\end{array}$

Now, using calculator calculate the angle .

  $\begin{array}{l}2\alpha \approx {11.31}^{\circ }\\ \Rightarrow \alpha \approx {5.65}^{\circ }\end{array}$

And,

  $\begin{array}{l}\sin \alpha \approx 0.10\\ \cos \alpha \approx 0.995\end{array}$

After rotation of coordinate axes, the original coordinates, $x$ and $y$ , are replaced by new coordinates $x^{\text{'}}$ and $y^{\text{'}}$ , such that,

  $\begin{array}{l}x=x^{\text{'}}\cos \alpha -y^{\text{'}}\sin \alpha \\ =0.995x^{\text{'}}-0.10y^{\text{'}}\\ y=x^{\text{'}}\sin \alpha +y^{\text{'}}\cos \alpha \\ =0.10x^{\text{'}}+0.995y^{\text{'}}\end{array}$

Substituting the above values of $x$ and $y$ in the given quadratic equation,

  $\begin{array}{l}2{\left(0.995x^{\text{'}}-0.10y^{\text{'}}\right)}^2-\left(0.995x^{\text{'}}-0.10y^{\text{'}}\right)\left(0.10x^{\text{'}}+0.995y^{\text{'}}\right)-3{\left(0.10x^{\text{'}}+0.995y^{\text{'}}\right)}^2\\ +3\left(0.995x^{\text{'}}-0.10y^{\text{'}}\right)-7=0\end{array}$

Simplify.

  $2.05x^{\text{'}}{}^2-3.05y^{\text{'}}{}^2+2.99x^{\text{'}}-0.30y^{\text{'}}-7=0$

Hence, the new equation is a hyperbola, having coefficients.

  $A^{\text{'}}=2.05,B^{\text{'}}=0.00,C^{\text{'}}=-3.05,D^{\text{'}}=2.99,E^{\text{'}}=-0.30,F^{\text{'}}=-7$

Conclusion:

The new equation is a hyperbola.

The angle is $\alpha \approx {5.65}^{\circ }$ .

Sine and cosine angle is,

  $\begin{array}{l}\sin \alpha \approx 0.10\\ \cos \alpha \approx 0.995\end{array}$