Chapter A5.3, Problem 32E

To determine

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

Answer to Problem 32E

The new equation forms parallel horizontal lines.

The angle is $\alpha \approx {18.43}^{\circ }$ .

Sine and cosine angle is,

  $\begin{array}{l}\sin \alpha \approx 0.32\\ \cos \alpha \approx 0.95\end{array}$

Explanation of Solution

Given:

  $2x^2-12xy+18y^2-49=0$

Calculation:

Comparing with the standard equation for a quadratic curve,

  $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$

Thus,

  $A=2,B=-12,C=18$

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

  $\begin{array}{l}\tan 2\alpha =\frac{B}{A-C}\\ =\frac{-12}{2-18}\\ =\frac{3}{4}\end{array}$

Now, using calculator calculate the angle ,

  $\begin{array}{l}2\alpha \approx {36.87}^{\circ }\\ \Rightarrow \alpha \approx {18.43}^{\circ }\end{array}$

And,

  $\begin{array}{l}\sin \alpha \approx 0.32\\ \cos \alpha \approx 0.95\end{array}$

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

  $\begin{array}{l}x=x^{\text{'}}\cos \alpha -y^{\text{'}}\sin \alpha \\ =0.95x^{\text{'}}-0.32y^{\text{'}}\\ y=x^{\text{'}}\sin \alpha +y^{\text{'}}\cos \alpha \\ =0.32x^{\text{'}}+0.95y^{\text{'}}\end{array}$

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

  $2{\left(0.95x^{\text{'}}-0.32y^{\text{'}}\right)}^2-12\left(0.95x^{\text{'}}-0.32y^{\text{'}}\right)\left(0.32x^{\text{'}}+0.95y^{\text{'}}\right)+18{\left(0.32x^{\text{'}}+0.95y^{\text{'}}\right)}^2-49=0$

Simplify

  $20.00y^{\text{'}}{}^2-49=0$

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

  $A^{\text{'}}=0.000,B^{\text{'}}=0.00,C^{\text{'}}=20.00,D^{\text{'}}=0,E^{\text{'}}=0,F^{\text{'}}=-49$

Conclusion:

The new equation forms parallel horizontal lines.

The angle is $\alpha \approx {18.43}^{\circ }$

Sine and cosine angle is,

  $\begin{array}{l}\sin \alpha \approx 0.32\\ \cos \alpha \approx 0.95\end{array}$