Chapter 11, Problem 56RE

a.

To determine

To Find whether graph of equation is parabola, ellipse or hyperbola from discriminant method.

Answer to Problem 56RE

The equation represent an ellipse.

Explanation of Solution

Given information:

Equation of the given conic is

  $\begin{array}{l}5x^2-6xy+5y^2-8x+8y-8=0\\ a=5\\ h=-3\\ b=4\\ g=-4\\ f=4\\ c=-8\\ △=\left|\begin{array}{ccc}5& -3& -4\\ 3& 5& 4\\ -4& 4& -8\end{array}\right|\\ =5\left(-40-16\right)+3\left(24+16\right)-4\left(-12+2\right)\\ =5\left(-56\right)+120-4\times 8\\ △=-280+120-32=-192\end{array}$

Concept Used:

For any 2^nd degree conic equation

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

If $△=0$ , it represents straight lines

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

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

Calculation:

So $△\ne 0$

So new

  $\begin{array}{l}{h}^2-4ab\Rightarrow 9-4\left(5\right)\left(5\right)<0\\ 9-100<0\\ -91<0\end{array}$

Interpretation:

So, the equation represents an ellipse.

b.

To determine

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

Answer to Problem 56RE

The equation is $2y^2+3x^2-8\sqrt{2}x-8=0$

Explanation of Solution

Given information:

Equation of curve

  $5x^2-6xy+5y^2-8x+8y-8=0$........ 3Concept Used:

To remove $xy$ from the equation

For example: $A{x}^2+C{y}^2+Bxy+Dx+Ey+F=\mathrm{0......}\left(1\right)$

to

  $A\text{'}x{\text{'}}^2+C\text{'}y{\text{'}}^2+D\text{'}x\text{'}+E\text{'}y\text{'}+F\text{'}=\mathrm{0......}\left(2\right)$

You need to rotate coordinate axis through angle $\theta$

  $\cot 2\theta =\frac{A-C}{B}$

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

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

Then replace them in the 1^st equation and get equation removed from $xy$ term.

Calculation:

  $\begin{array}{l}A=5\\ B=-6\\ C=5\\ \cot 2\theta =\frac{A-C}{B}=\frac{5-5}{-6}\text{=0 }\\ 2\theta =0\Rightarrow 2\theta =\frac{\pi }{2}\Rightarrow \theta =\frac{\pi }{4}\\ x\text{'}=\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}},y\text{'}=\frac{-x}{\sqrt{2}}+\frac{y}{\sqrt{2}}\end{array}$

Now put in equation 3

  $\begin{array}{l}\frac{5}{2}{\left(x+y\right)}^2+\frac{-6}{2}\left(x+y\right)\left(y-x\right)+\frac{5}{2}{\left(y-x\right)}^2\\ -\frac{8}{\sqrt{2}}\left(x+y\right)+\frac{8}{\sqrt{2}}\left(y-x\right)-8=0\\ 5y^2+3x^2-3y^2-8\sqrt{2}-8=0\\ 2y^2+3x^2-8\sqrt{2}x-8=0\end{array}$

Interpretation:

  $xy$ term has been from conic equation using rotation of axis.and obtained values are $2y^2+3x^2-8\sqrt{2}x-8=0$

c.

To determine

To draw: The graph of the given curve.

Explanation of Solution

Given information:

The equation $5x^2-6xy+5y^2-8x-8x+8y-8=0$

  

Interpretation :

The graph of the given curve has been obtained