Chapter 9, Problem 61RE

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 61RE

The equation in standard form is a $\frac{{(x^{\text{'}})}^2}{3}-\frac{{(y^{\text{'}})}^2}{2}=1$$5x^2-2xy+5y^2-12=0$

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Explanation of Solution

Given:

It is given in the question that the equation is .

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 $5x^2-2xy+5y^2-12=0$ .

Now, Find a,b and c $\Rightarrow a=5,b=-2,c=5$ .

Now, find theta,

  $\begin{array}{l}\cot \theta =\frac{A-C}{B}\\ \cot \theta =\frac{5-5}{-2}\\ \cot \theta =0\\ 2\theta =\frac{\pi }{2}\\ \theta =\frac{\pi }{4}\end{array}$

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

  $\begin{array}{l}5x^2-2xy+5y^2-12=0\\ $ 5{(\frac{x^{\text{'}}-y\text{'}}{\sqrt{2}})}^2-2(\frac{x^{\text{'}}-y\text{'}}{\sqrt{2}})(\frac{x^{\text{'}}-y\text{'}}{\sqrt{2}})+5{(\frac{x^{\text{'}}-y\text{'}}{\sqrt{2}})}^2-12=0$$ 4{(x^{\text{'}})}^2+6{(y\text{'})}^2-12=0$$ 4{(x^{\text{'}})}^2+6{(y\text{'})}^2=12$\frac{{(x^{\text{'}})}^2}{3}-\frac{{(y^{\text{'}})}^2}{2}=1\end{array}$

Now, the equation of the graph is given below:

  

Conclusion:

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