Chapter 10, Problem 25RE

In Problems 24-26, rotate the axes so that the new equation contains no $xy\text{-term}$. Analyze and graph the new equation.

$6x^2+4xy+9y^2-20=0$

To determine

a. The new equation that contains no $xy$ term.

Answer to Problem 25RE

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

Explanation of Solution

Given:

$6x^2+4xy+9y^2-20=0$

Calculation:

Consider the given equation $6x^2+4xy+9y^2-20=0$.

$\text{Cot}2\theta =\frac{A-B}{C},\cot 2\theta =\frac{6-9}{4}=-\frac{3}{4};\cos 2\theta =-\frac{3}{5};\sin \theta =\sqrt{1-{\left(-\frac{3}{5}\right)}^2}=\frac{2}{\sqrt{5}},\cos \theta =\frac{1}{\sqrt{5}};\theta =63.4$

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

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

$6{(\frac{1}{\sqrt{5}}\left(x^{\prime }-2y^{\prime }\right))}^2+4*\frac{1}{\sqrt{5}}\left(x^{\prime }-2y^{\prime }\right)*\frac{\text{1}}{\sqrt{\text{5}}}\left(2x^{\text{'}}+y^{\prime }\right)+9{\left(\frac{1}{\sqrt{5}}\left(2x^{\text{'}}+y^{\prime }\right)\right)}^2-20=0$

$\frac{6}{5}({x^{\prime }}^2+y^{\prime }²-4x^{\prime }{y}^{\prime })+\frac{4}{5}\left(2{x^{\prime }}^2-2y^{\prime }²-3x^{\prime }{y}^{\prime }\right)+\frac{9}{5}(4{x^{\prime }}^2+y^{\prime }²+4x^{\prime }{y}^{\prime })-20=0$

$6{x^{\prime }}^2+24y^{\prime }²-24x^{\prime }{y}^{\prime }+8{x^{\prime }}^2-8y^{\prime }²-12x^{\prime }{y}^{\prime }+36{x^{\prime }}^2+9y^{\prime }²+36x^{\prime }{y}^{\prime }-100=0+50{x^{\prime }}^2-25y^{\prime }²=100$

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

It is an ellipse with center at the origin; Major axis is the $y^{\prime }\text{axis}$, vertices $\left(0,\pm 2\right)$, foci at $\left(x^{\prime },<mtext> </mtext>y^{\prime }\right)=\left(0,\pm \surd 2\right)$.

To determine

To graph:

b. $\frac{{x^{\prime }}^2}{2}-\frac{{y^{\prime }}^2}{4}=1$ in $x^{\prime }{y}^{\prime }\text{-plane}$.

Answer to Problem 25RE

Explanation of Solution

Given:

$6x^2+4xy+9y^2-20=0$

Graph:

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

The graph of $6x^2+4xy+9y^2-20=0$ which is also the graph of $\frac{{x^{\prime }}^2}{2}+\frac{{y^{\prime }}^2}{4}=1$ in $x^{\prime }{y}^{\prime }\text{-plane}$.

It is an ellipse with center at the origin; Major axis is the $y^{\prime }\text{axis}$, vertices $\left(0,\pm 2\right)$, foci at $\left(x^{\prime },<mtext> </mtext>y^{\prime }\right)=\left(0,\pm \surd 2\right)$.