Chapter 9, Problem 60RE

To determine

Find the equation in standard form by rotating the axes to eliminate the xy-term in the equation. Also draw the graph of the equation.

Answer to Problem 60RE

The equation in standard form is $\frac{{(y^{\text{'}})}^2}{-3}-\frac{-{(x^{\text{'}})}^2}{(-9)}=1$ ..

Explanation of Solution

Given:

It is given in the question that the equation is $x^2-4xy+y^2+9=0$ .

Concept Used:

In this, use the concept of identifying a, b and c from the equation and also use the formula $\cot \theta =\frac{A-C}{B}$ .

Calculation:

Here, the equation is $x^2-4xy+y^2+9=0$ .

Now, Find a,b and c $\Rightarrow a=1,b=-4,c=1$ .

Now, find theta,

  $\begin{array}{l}\cot 2\theta =\frac{A-C}{B}\\ \cot 2\theta =\frac{0-0}{-4}\\ \cot 2\theta =0\\ 2\theta =\frac{\pi }{2}\\ \theta =\frac{\pi }{4}\end{array}$

Again, substitute the equation for x and y into the given equation to eliminate to xy term.

  $\begin{array}{l}{x}^2-4xy+y^2+9=0\\ $ {(\frac{x^{\text{'}}-y^{\text{'}}}{\sqrt{2}})}^2-4(\frac{x^{\text{'}}-y^{\text{'}}}{\sqrt{2}})(\frac{x^{\text{'}}+y^{\text{'}}}{\sqrt{2}})+{(\frac{x^{\text{'}}+y^{\text{'}}}{\sqrt{2}})}^2=-9$$ [\frac{1}{2}{(x^{\text{'}})}^2-x^{\text{'}}{y}^{\text{'}}+\frac{1}{2}{(y^{\text{'}})}^2]-2{(x^{\text{'}})}^2+2{(y^{\text{'}})}^2+[\frac{1}{2}{(x^{\text{'}})}^2+x^{\text{'}}{y}^{\text{'}}+\frac{1}{2}{(y^{\text{'}})}^2]=-9$$ -{(x^{\text{'}})}^2+3{(y^{\text{'}})}^2=-9$\frac{{(y^{\text{'}})}^2}{-3}-\frac{-{(x^{\text{'}})}^2}{(-9)}=1\end{array}$

Now, the equation of the graph is given below:

  

Conclusion:

The equation is $\frac{{(y^{\text{'}})}^2}{-3}-\frac{-{(x^{\text{'}})}^2}{(-9)}=1$ .