Chapter 11, Problem 57RE

a.

To determine

To find: Whether the graph of the equation is a parabola, an ellipse or a hyperbola.

Answer to Problem 57RE

The graph of the equation is an ellipse

Explanation of Solution

Given information:

The equation of the graph is ${\text{7x}}^2-6\sqrt{3}x+13y^2-4\sqrt{3}x-4y=0$

Concept used:

The general equation of the form

  $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$

The discremanant is

  $\Delta =\left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\right|=abc+2fgh-a{f}^2-b{g}^2-c{h}^2$

If $\Delta =0$ then it represents degenerate conic and $\Delta \ne 0$ then it generate non − degenerate conic

If $\Delta \ne 0$

  $h^2-ab>0$ represents hyperbola

  $h^2-ab=0$ represents parabola.

  $h^2-ab<0$ represents ellipse

Calculation:

The given equation is ${\text{7x}}^2-6\sqrt{3}x+13y^2-4\sqrt{3}x-4y=0$

Comparing with general equation

  $\begin{array}{l}a=7\\ h=-3\sqrt{3}\\ b=13\\ g=-2\sqrt{3}\\ f=-2\\ c=0\\ \Delta =\left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\right|\ne 0\\ h^2-ab={\left(-3\sqrt{3}\right)}^2-\left(7\right)\left(13\right)=-64\end{array}$

Conclusion:

Since $h^2-ab<0$

The equation represents an ellipse.

b.

To determine

To eliminate: $xy$ term in the given equation.

Answer to Problem 57RE

Required equation after eliminating $xy$ term is

  $4x^2+16y^2-9x=0$

Explanation of Solution

Given information:

The equation of the graph is ${\text{7x}}^2-6\sqrt{3}x+13y^2-4\sqrt{3}x-4y=0$

Concept used:

Let $\theta$ be the angle of rotation of axis then

  $\mathrm{Tan}2\theta =\left(\frac{2h}{a-b}\right)$

Using this find $\sin \theta$ and $\cos \theta$

The new axis will be

  $\begin{array}{l}x=x\text{'}\cos \theta -y\text{'}\sin \theta \\ y=x\text{'}\sin \theta +y\text{'}\cos \theta \end{array}$

Substitute x & y in the given equation t obtain the required equation.

Calculation:

Comparing the given equation

  ${\text{7x}}^2-6\sqrt{3}x+13y^2-4\sqrt{3}x-4y=0$

with the general equation

  $\begin{array}{l}a=7\\ b=13\\ c=0\\ g=-2\sqrt{3}\\ h=-3\sqrt{3}\\ f=-2\end{array}$

Substitute the values find $\sin \theta$ and $\cos \theta$

  $\begin{array}{l}\mathrm{Tan}2\theta =\frac{2h}{a-b}=\frac{6\sqrt{3}}{6}=\sqrt{3}\\ \Rightarrow \theta =\frac{1}{2}{\mathrm{Tan}}^{-1}\left(\sqrt{3}\right)=\frac{1}{2}\frac{\pi }{3}=\frac{\pi }{6}\\ \sin \theta =\sin \frac{\pi }{6}=\frac{1}{2}\&\cos \theta =\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}\end{array}$

The new axis are

  $\begin{array}{l}x=x\text{'}\cos \theta -y\text{'}\sin \theta =\frac{\sqrt{3}}{2}x\text{'}+\frac{y\text{'}}{2}\\ y=x\text{'}\sin \theta +y\text{'}\cos \theta =\frac{x\text{'}}{2}+\frac{\sqrt{3}}{2}y\text{'}\end{array}$

Substitute a and y in given equation

  $\begin{array}{l}{\text{7x}}^2-6\sqrt{3}x+13y^2-4\sqrt{3}x-4y=0\\ \Rightarrow 7{\left(\frac{\sqrt{3}}{2}x\text{'}-\frac{y\text{'}}{2}\right)}^2-6\sqrt{3}\left(\frac{\sqrt{3}}{2}x\text{'}-\frac{y\text{'}}{2}\right)\left(\frac{x\text{'}}{2}+\frac{\sqrt{3}}{2}y\text{'}\right)+13\left(\frac{x\text{'}}{2}+\frac{\sqrt{3}}{2}y\text{'}\right)-4\sqrt{3}\left(\frac{\sqrt{3}}{2}x\text{'}-\frac{y\text{'}}{2}\right)-4\left(\frac{x\text{'}}{2}+\frac{\sqrt{3}}{2}y\text{'}\right)=0\end{array}$

Simplifying the above equation

  $4{\left(x\text{'}\right)}^2+16{\left(y\text{'}\right)}^2-9\left(x\text{'}\right)=0$

Conclusion:

Therefore the required equation after eliminating the $xy$ term is

  $4x^2+16y^2-9x=0$

c.

To determine

To sketch: The graph of the given equation

Answer to Problem 57RE

From the graph it can be observed that the given equation represents an ellipse.

Explanation of Solution

Given information:

Given equation of graph is ${\text{7x}}^2-6\sqrt{3}x+13y^2-4\sqrt{3}x-4y=0$

Graph:

The graph is obtained as:

  

Interpretation:

From the above graph it can be observed that the given equation represents an ellipse.