Chapter 10.5, Problem 27AYU

To determine

The appropriate rotation formula to use so that new equation contains no $xy$ term: $4x^2-4xy+y^2-8\sqrt{5x}-16\surd 5y=0$.

Answer to Problem 27AYU

$x=\sqrt{\frac{1}{5}}\left(x^{\prime }-2y^{\prime }\right)$, $y=\sqrt{\frac{1}{5}}\left(2x^{\prime }+y^{\prime }\right)$

Explanation of Solution

Given:

$4x^2-4xy+y^2-8\sqrt{5x}-16\surd 5y=0$

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 $4x^2-4xy+y^2-8\sqrt{5x}-16\surd 5y=0$.

$A=4, B=-4 \text{and} C=1$;

$\text{Cot}\left(2\theta \right)=\frac{A-C}{B};\frac{4-1}{-4}=\frac{3}{-4}=-\frac{3}{4};\cos 2\theta =-\frac{3}{5}$

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

$\text{sin}\theta =\surd \frac{1+3/5}{2}$; $\text{cos}\theta =\surd \frac{1-3/5}{2}$;

$\text{sin}\theta =2\sqrt{\frac{1}{5}}, \text{cos}\theta =\sqrt{\frac{1}{5}}$

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

$x=x^{\prime }\sqrt{\frac{1}{5}}-y^{\prime }2\sqrt{\frac{1}{5}}$

$x=\sqrt{\frac{1}{5}}\left(x^{\prime }-2y^{\prime }\right)$

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

$y=2x^{\prime }\sqrt{\frac{1}{5}}+y^{\prime }\sqrt{\frac{1}{5}}$

$y=\sqrt{\frac{1}{5}}\left(2x^{\prime }+y^{\prime }\right)$