Chapter 9, Problem 62RE

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 62RE

The equation in standard form is ${(x^{\text{'}}-\frac{1}{8})}^2=-2(y^{\text{'}}-\frac{1}{128})$ .

Explanation of Solution

Given:

It is given in the question that the equation is $4x^2+8xy+4y^2+7\sqrt{2}x+9\sqrt{2}y=0$ .

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 $4x^2+8xy+4y^2+7\sqrt{2}x+9\sqrt{2}y=0$ .

Now, Find a,b and c $\Rightarrow a=4,b=8,c=4$ .

Now, find theta,

  $\begin{array}{l}\cot 2\theta =\frac{A-C}{B}\\ \cot 2\theta =\frac{4-4}{8}\\ \cot 2\theta =0\\ 2\theta =\frac{\pi }{2}\\ \theta =\frac{\pi }{4}\end{array}$

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

  $\begin{array}{l}4x^2+8xy+4y^2+7\sqrt{2}x+9\sqrt{2}y=0\\ $ 4{(\frac{x^{\text{'}}-y\text{'}}{\sqrt{2}})}^2+8(\frac{x^{\text{'}}-y\text{'}}{\sqrt{2}})(\frac{x^{\text{'}}+y\text{'}}{\sqrt{2}})+5{(\frac{x^{\text{'}}+y\text{'}}{\sqrt{2}})}^2+7\sqrt{2}(\frac{x^{\text{'}}-y\text{'}}{\sqrt{2}})+9\sqrt{2}(\frac{x^{\text{'}}+y\text{'}}{\sqrt{2}})=0$$ 8{(x^{\text{'}})}^2+2(x\text{'})+16y^{\text{'}}=0$$ {(x^{\text{'}}-\frac{1}{8})}^2=-2(y^{\text{'}}-\frac{1}{128})$\end{array}$

Now, the equation of the graph is given below:

  

Conclusion:

The equation is ${(x^{\text{'}}-\frac{1}{8})}^2=-2(y^{\text{'}}-\frac{1}{128})$ .