Chapter A5.3, Problem 31E

To determine

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

Answer to Problem 31E

The new equation forms parallel horizontal lines.

The angle is $\alpha \approx {26.57}^{\circ }$ .

Sine and cosine angle is,

  $\begin{array}{l}\sin \alpha \approx 0.45\\ \cos \alpha \approx 0.89\end{array}$

Explanation of Solution

Given:

  $x^2-4xy+4y^2-5=0$

Calculation:

Comparing with the standard equation for a quadratic curve,

  $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$

Thus,

  $A=1,B=-4,C=4$

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

  $\begin{array}{l}\tan 2\alpha =\frac{B}{A-C}\\ =\frac{-4}{1-4}\\ =\frac{4}{3}\end{array}$

Now, using calculator calculate the angle,

  $\begin{array}{l}2\alpha \approx {53.13}^{\circ }\\ \Rightarrow \alpha \approx {26.57}^{\circ }\end{array}$

And,

  $\begin{array}{l}\sin \alpha \approx 0.45\\ \cos \alpha \approx 0.89\end{array}$

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

  $\begin{array}{l}x=x^{\text{'}}\cos \alpha -y^{\text{'}}\sin \alpha \\ =0.89x^{\text{'}}-0.45y^{\text{'}}\\ y=x^{\text{'}}\sin \alpha +y^{\text{'}}\cos \alpha \\ =0.45x^{\text{'}}+0.89y^{\text{'}}\end{array}$

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

  ${\left(0.89x^{\text{'}}-0.45y^{\text{'}}\right)}^2-4\left(0.89x^{\text{'}}-0.45y^{\text{'}}\right)\left(0.45x^{\text{'}}+0.89y^{\text{'}}\right)+4{\left(0.45x^{\text{'}}+0.89y^{\text{'}}\right)}^2-5=0$

Simplify.

  $5.00y^{\text{'}}{}^2-5=0$

Hence, the new equation forms parallel horizontal lines, having coefficients.

  $A^{\text{'}}=0.050,B^{\text{'}}=0.00,C^{\text{'}}=5.00,D^{\text{'}}=0,E^{\text{'}}=0,F^{\text{'}}=-5$

Conclusion:

The new equation forms parallel horizontal lines.

The angle is $\alpha \approx {26.57}^{\circ }$ .

Sine and cosine angle is,

  $\begin{array}{l}\sin \alpha \approx 0.45\\ \cos \alpha \approx 0.89\end{array}$