Chapter 11, Problem 55RE

a.

To determine

To find: whether graph of equation is parabola, ellipse or hyperbola with the help of discriminant method.

Answer to Problem 55RE

The curve is a rectangular hyperbola.

Explanation of Solution

Given information:

The equation is $x^2+4xy+y^2-1=0$

Concept Used:

For any 2^nd degree conic equation

  $\begin{array}{l}a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0\\ △=\left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\right|\end{array}$

Calculation:

  $\begin{array}{l}a=1\text{ g = 0}\\ h=2\text{ f = 0}\\ c=-1\\ b=1\\ △=\left|\begin{array}{ccc}1& 2& 0\\ 2& 1& 0\\ 0& 0& -1\end{array}\right|=1\left(-1-0\right)-2\left(-2-0\right)\\ =-1+4\ne 0\\ h^2-4ab\\ 1-4\left(1\right)\left(-1\right)=5>0\end{array}$

Also, $a+c=0$

So, the equation of the conic is a rectangular hyperbola.

Conclusion:

If $△=0$ , it represents straight lines

If $△\ne 0$ , it represents the conics

  $\begin{array}{l}A{x}^2+C{y}^2+Bxy+Dx+Ey+F=0\\ h^2-4ab>0\text{ , it represents hyperbola}\\ h^2-4ab=0\text{ , parabola}\\ h^2-4ab<0\text{ , ellipse}\end{array}$

b.

To determine

To remove: $xy$ term from conic equation using rotation of axis.

Answer to Problem 55RE

The equation eliminated from $xy$ is $3y^2-2x^2-1=0$

Explanation of Solution

Given information:

The equation of the curve

  $x^2+4xy+y^2-1=0$

Concept Used:

To remove $xy$ from rotation of axis

For example: $A{x}^2+C{y}^2+Bxy+Dx+Ey+F=0$-equation1

The given equation eeds to be converted to

  $A\text{'}x{\text{'}}^2+C\text{'}y{\text{'}}^2+D\text{'}x\text{'}+E\text{'}y\text{'}+F\text{'}=0$

-equation 2

Now, rotate coordinate axis through angle $\theta$ such that

  $\cot 2\theta =\frac{A-C}{B}$

Then replace $x\text{'}=x\cos \theta +y\sin \theta$

  $y\text{'}=-x\sin \theta +y\cos \theta$

in the second equation and find the equation where $xy$ has been eliminated.

Calculation:

  $\begin{array}{l}\cot 2\theta =\frac{A-C}{B}=0A=1,C=1,B=4\\ 2\theta =\frac{\pi }{2}\Rightarrow \theta =\frac{\pi }{4}\\ x\text{'}=\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}},y\text{'}=\frac{-x}{\sqrt{2}}+\frac{y}{\sqrt{2}}\end{array}$

Now put in equation 2

  $\begin{array}{l}{\left(\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}\right)}^2+4\left(\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}\right)\left(\frac{-x}{\sqrt{2}}+\frac{y}{\sqrt{2}}\right)+{\left(\frac{-x}{\sqrt{2}}+\frac{y}{\sqrt{2}}\right)}^2\\ \frac{1}{2}\left(x^2+y^2+2xy\right)+4\frac{\left(-x^2\right)}{2}+\frac{4xy}{2}-\frac{4xy}{2}+\frac{4y^2}{2}+\frac{y^2}{2}-\frac{x^2}{2}-2xy-1=0\\ \frac{y^2}{2}-2x^2+2y^2+\frac{y^2}{2}-1=0\\ 3y^2-2x^2-1=0\end{array}$

Conclusion:

The $xy$ term is removed from conic equation using rotation of axis which is:

  $3y^2-2x^2-1=0$

c.

To determine

To sketch: The graph of given curve.

Answer to Problem 55RE

From the graph it has been observed that the lines intersect at origin $x=0$

Explanation of Solution

Given information:

  $x^2+4xy+y^2=1$

  

Interpretation:

From the above graph it has been observed that the lines intersect at origin $x=0$