Chapter 10.5, Problem 32AYU

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.

$x^2-4xy+y^2-3=0$

To determine

a.

The new equation that contains no $xy$ term.

Answer to Problem 32AYU

${y^{\prime }}^2-\frac{{x^{\prime }}^2}{3}=1$

Explanation of Solution

Given:

$x^2-4xy+y^2-3=0$

Calculation:

Consider the given equation $x^2-4xy+y^2-3=0$.

Using problem 22, 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,

${\left(\frac{1}{\sqrt{2}}\left(x^{\prime }-y^{\prime }\right)\right)}^2-4*\frac{1}{\sqrt{2}}\left(x^{\prime }-y^{\prime }\right)\text{*}\frac{1}{\sqrt{2}}\left(x^{\prime }+y^{\prime }\right)+{\left(\frac{1}{\sqrt{2}}\left(x^{\prime }+y^{\prime }\right)\right)}^{\text{2}}-3=0$

$\frac{1}{2}({x^{\prime }}^2+{y^{\prime }}^2-2x^{\prime }{y}^{\prime })-2\left({x^{\prime }}^2-{y^{\prime }}^2\right)+\frac{1}{2}({x^{\prime }}^2+{y^{\prime }}^2+2x^{\prime }{y}^{\prime })-3=0$

$-{x^{\prime }}^2+3{y^{\prime }}^2=3$

${y^{\prime }}^2-\frac{{x^{\prime }}^2}{3}=1$

It is hyperbola with center at the origin; transverse axis is the $y^{\prime }\text{-axis}$, vertices $\left(0,\pm 1\right)$.

To determine

b.

To graph: ${y^{\prime }}^2-\frac{{x^{\prime }}^2}{3}=1$ in $x^{\prime }{y}^{\prime }\text{-plane}$.

Answer to Problem 32AYU

Explanation of Solution

Given:

$x^2-4xy+y^2-3=0$

Graph:

The graph of ${y^{\prime }}^2-\frac{{x^{\prime }}^2}{3}=1$ in $x^{\prime }{y}^{\prime }\text{-plane}$.

The graph of $x^2-4xy+y^2-3=0$ which is also the graph of ${y^{\prime }}^2-\frac{{x^{\prime }}^2}{3}=1$ in $x^{\prime }{y}^{\prime }\text{-plane}$.

It is hyperbola with center at origin; transverse axis is the $y^{\prime }\text{-axis}$, vertices $\left(0,\pm 1\right)$.