Chapter 8.4, Problem 8E

Solution

To find: the equation $\left(u,v\right)$ form of $5x^2+6xy+5y^2=16$ using the substitutions $x=\frac{u}{\sqrt{2}}-\frac{v}{\sqrt{2}}$ and $y=\frac{u}{\sqrt{2}}+\frac{v}{\sqrt{2}}$ .

Thus, the required equation in $\left(u,v\right)$ form is $4u^2+v^2=8$ .

Given information: The given equations are $5x^2+6xy+5y^2=16$ , $x=\frac{u}{\sqrt{2}}-\frac{v}{\sqrt{2}}$ , and $y=\frac{u}{\sqrt{2}}+\frac{v}{\sqrt{2}}$ .

Formula Used:

The algebraic identities are $\left(a-b\right)\left(a+b\right)=a^2-b^2$ , ${\left(a+b\right)}^2=a^2+2ab+b^2$ , and ${\left(a-b\right)}^2=a^2-2ab+b^2$ .

Calculation:

Substitute the given values $x=\frac{u}{\sqrt{2}}-\frac{v}{\sqrt{2}}$ and $y=\frac{u}{\sqrt{2}}+\frac{v}{\sqrt{2}}$ into the given equation, $5x^2+6xy+5y^2=16$ and simplify using the identities, $\left(a-b\right)\left(a+b\right)=a^2-b^2$ , ${\left(a+b\right)}^2=a^2+2ab+b^2$ , and ${\left(a-b\right)}^2=a^2-2ab+b^2$ .

  $\begin{array}{c}5x^2+6xy+5y^2=16\\ 5{\left(\frac{u}{\sqrt{2}}-\frac{v}{\sqrt{2}}\right)}^2+6\left(\frac{u}{\sqrt{2}}-\frac{v}{\sqrt{2}}\right)\left(\frac{u}{\sqrt{2}}+\frac{v}{\sqrt{2}}\right)+5{\left(\frac{u}{\sqrt{2}}+\frac{v}{\sqrt{2}}\right)}^2=16\\ 5\left(\frac{u^2}{2}-uv+\frac{v^2}{2}\right)+6\left(\frac{u^2}{2}-\frac{v^2}{2}\right)+5\left(\frac{u^2}{2}+uv+\frac{v^2}{2}\right)=16\\ \frac{5u^2}{2}-5uv+\frac{5v^2}{2}+3u^2-3v^2+\frac{5u^2}{2}+5uv+\frac{5v^2}{2}=16\end{array}$

Combine the like terms and simplify the obtained equation.

  $\begin{array}{c}5u^2+3u^2+5v^2-3v^2=16\\ 8u^2+2v^2=16\\ 4u^2+v^2=8\end{array}$

Thus, the required equation in $\left(u,v\right)$ form is $u^2+4v^2=8$ .