Chapter 10.5, Problem 33AYU

In Problems 31-42, rotate the axes so that the new equation contains no $xy$-term. Analyze and graph the new equation. Refer to Problems 21-30 for Problems 31-40.

$5x^2+6xy+5y^2-8=0$

To determine

a.

The new equation that contains no $xy$ term.

Answer to Problem 33AYU

${x^{\prime }}^2+\frac{{y^{\prime }}^2}{4}=1$

Explanation of Solution

Given:

$5x^2+6xy+5y^2-8=0$

Calculation:

Consider the given equation $5x^2+6xy+5y^2-8=0$.

Using problem 23, we find that $\theta =45°$, $x=\frac{1}{\sqrt{2}}\left(x^{\prime }-y^{\prime }\right)$, $y=\frac{1}{\sqrt{2}}\left(x^{\prime }+y^{\prime }\right)$.

Substituting $x$, $y$ in the above equation we get,

$5{\left(\frac{1}{\sqrt{2}}\left(x^{\prime }-y^{\prime }\right)\right)}^2+6*\frac{1}{\sqrt{2}}\left(x^{\prime }-y^{\prime }\right)\text{*}\frac{1}{\sqrt{2}}\left(x^{\prime }+y^{\prime }\right)+5{\left(\frac{1}{\sqrt{2}}\left(x^{\prime }+y^{\prime }\right)\right)}^2-8=0$

$\frac{5}{2}({x^{\prime }}^2+{y^{\prime }}^2-2x^{\prime }{y}^{\prime })+3\left({x^{\prime }}^2-{y^{\prime }}^2\right)+\frac{5}{2}({x^{\prime }}^2+{y^{\prime }}^2+2x^{\prime }{y}^{\prime })-8=0$

$8{x^{\prime }}^2+2{y^{\prime }}^2=8$

${x^{\prime }}^2+\frac{{y^{\prime }}^2}{4}=1$

It is an ellipse with center at the origin; major axis is the $y^{\prime }\text{-axis}$, vertices $\left(0,\pm 2\right)$.

To determine

b.

To graph: ${x^{\prime }}^2+\frac{{y^{\prime }}^2}{4}=1$ in $x^{\prime }{y}^{\prime }\text{-plane}$.

Answer to Problem 33AYU

Explanation of Solution

Given:

$5x^2+6xy+5y^2-8=0$

Graph:

The graph of ${x^{\prime }}^2+\frac{{y^{\prime }}^2}{4}=1$ in $x^{\prime }{y}^{\prime }\text{-plane}$.

The graph of $5x^2+6xy+5y^2-8=0$ which is also the graph of ${x^{\prime }}^2+\frac{{y^{\prime }}^2}{4}=1$ in $x^{\prime }{y}^{\prime }\text{-plane}$.

It is ellipse with center at origin; major axis is the $y^{\prime }\text{-axis}$, vertices $\left(0,\pm 2\right)$.