Chapter 10.5, Problem 30AYU

To determine

The appropriate rotation formula to use so that new equation contains no $xy$ term: $34x^2-24xy+41y^2-25=0$.

Answer to Problem 30AYU

$x=\frac{1}{5}\left(4x^{\prime }-3y^{\prime }\right)$, $y=\frac{1}{5}(3x^{\prime }+4y^{\prime })$

Explanation of Solution

Given:

$34x^2-24xy+41y^2-25=0$

Formula used:

To transform $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0, B\ne 0$ into an equation in $x^{\prime }$ and $y^{\prime }$ without an $x^{\prime }{y}^{\prime }$-term, rotate the axes through an angle $\theta$ that satisfies the equation $\text{Cot}\left(2\theta \right)=\frac{A-C}{B}$.

$x = x^{\prime }\text{cos}\theta - y^{\prime }\text{sin}\theta ; y= x^{\prime }\text{sin}\theta + y^{\prime }\text{cos}\theta$

Calculation:

Consider the given equation $34x^2-24xy+41y^2-25=0$.

$A=34, B=-24 \text{and} C=41$;

$\text{Cot}\left(2\theta \right)=\frac{A-C}{B};\frac{34-41}{-24}=\frac{-7}{-24}=\frac{7}{24};\cos 2\theta =\frac{7}{25}$

$\text{sin}\theta =\surd \frac{1-\text{cos}2\theta }{2}$; $\text{cos}\theta =\surd \frac{1+\text{cos}2\theta }{2}$;

$\text{sin}\theta =\sqrt{\frac{1-\frac{7}{25}}{2}, }\text{cos}\theta =\sqrt{\frac{1+\frac{7}{25}}{2}}$

$\text{sin}\theta =\frac{3}{5}, \text{cos}\theta =\frac{4}{5}$

$x = x^{\prime }\text{cos}\theta - y^{\prime }\text{sin}\theta$

$x=\frac{4}{5}{x}^{\prime }-\frac{3}{5}{y}^{\prime }$

$x=\frac{1}{5}\left(4x^{\prime }-3y^{\prime }\right)$

$y = x^{\prime }\text{sin}\theta + y^{\prime }\text{cos}\theta$

$y=\frac{3}{5}{x}^{\prime }+\frac{4}{5}{y}^{\prime }$

$y=\frac{1}{5}(3x^{\prime }+4x^{\prime })$