Chapter 11.5, Problem 26E

(a)

To determine

To check: Whether the graph of the equation $9x^2-24xy+16y^2=100\left(x-y-1\right)$ is a parabola, an ellipse or a hyperbola with the help of discriminant.

Answer to Problem 26E

The graph of the equation $9x^2-24xy+16y^2=100\left(x-y-1\right)$ is a parabola.

Explanation of Solution

Given information:

The equation of the conic is $9x^2-24xy+16y^2=100\left(x-y-1\right)$ .

Calculation:

Simplify the equation of the conic.

  $\begin{array}{c}9x^2-24xy+16y^2=100\left(x-y-1\right)\\ 9x^2-24xy+16y^2=100x-100y-100\\ 9x^2-24xy+16y^2-100x+100y+100=0\end{array}$

Compare the given equation of conic with general equation $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ .

This gives the value of $A$ , $B$ , $C$ , $D$ , $E$ and $F$ equal to $9$ , $-24$ , $16$ , $-100$ , $100$ and $100$ respectively.

Now check the value of discriminant $B^2-4AC$ .

Substitute $9$ , $-24$ , $16$ for $A$ , $B$ , $C$ respectively in the discriminant.

  $\begin{array}{c}{B}^2-4AC={\left(-24\right)}^2-4\times 9\times 16\\ =576-576\\ =0\end{array}$

The value of discriminant is equal to zero.

Therefore, the graph of the equation $9x^2-24xy+16y^2=100\left(x-y-1\right)$ is a parabola.

(b)

To determine

To find: The equation of the conic $9x^2-24xy+16y^2=100\left(x-y-1\right)$ with the elimination of $xy$ term with rotation of axes.

Answer to Problem 26E

The equation of the conic with the elimination of $xy$ term is equal to ${\left(5Y+14\right)}^2=20X+96$ .

Explanation of Solution

Given information:

The equation of the conic is $9x^2-24xy+16y^2=100\left(x-y-1\right)$ .

Calculation:

To eliminate the $xy$ term from the given equation of conic take the angle such that $\cot 2\varphi =\frac{A-C}{B}$ .

Substitute $9$ , $-24$ , $16$ for $A$ , $B$ , $C$ in the equation for angle of rotation.

  $\begin{array}{c}\cot 2\varphi =\frac{A-C}{B}\\ =\frac{9-16}{-24}\\ =\frac{-7}{-24}\\ =\frac{7}{24}\end{array}$

The formula for the trigonometric function is the ratio of the adjacent to the opposite.

Use the Pythagoras theorem for the hypotenuse.

  $\begin{array}{c}{H}^2=P^2+B^2\\ H^2={\left(24\right)}^2+{\left(7\right)}^2\\ H=\sqrt{576+49}\\ H=25\end{array}$

The value of the trigonometric function cosine is the ratio of the adjacent to the hypotenuse.

So, the value of $\cos 2\varphi$ is equal to $\frac{7}{25}$ .

Now use the half angle formulas $\cos \varphi =\sqrt{\frac{1+\cos 2\varphi }{2}}$ and $\sin \varphi =\sqrt{\frac{1-\cos 2\varphi }{2}}$ for the value of $\cos \varphi$ and $\sin \varphi$ respectively.

  $\begin{array}{c}\cos \varphi =\sqrt{\frac{1+\cos 2\varphi }{2}}\\ =\sqrt{\frac{1+\frac{7}{25}}{2}}\\ =\sqrt{\frac{\frac{25+7}{25}}{2}}\\ =\sqrt{\frac{32}{2\times 25}}\end{array}$

Further simplify,

  $\begin{array}{c}\cos \varphi =\sqrt{\frac{32}{2\times 25}}\\ =\sqrt{\frac{16}{25}}\\ =\frac{4}{5}\end{array}$

Now simplify for the value of $\sin \varphi$ .

  $\begin{array}{c}\sin \varphi =\sqrt{\frac{1-\cos 2\varphi }{2}}\\ =\sqrt{\frac{1-\frac{7}{25}}{2}}\\ =\sqrt{\frac{\frac{25-7}{25}}{2}}\\ =\sqrt{\frac{18}{2\times 25}}\end{array}$

Further simplify,

  $\begin{array}{c}\sin \varphi =\sqrt{\frac{18}{2\times 25}}\\ =\sqrt{\frac{9}{25}}\\ =\frac{3}{5}\end{array}$

Now use the transformation $x=\cos \varphi X-\sin \varphi Y$ and $y=\sin \varphi X+\cos \varphi Y$ .

Substitute the value of $\sin \varphi$ and $\cos \varphi$ , the transformation changes to $x=\frac{4}{5}X-\frac{3}{5}Y$ and $y=\frac{3}{5}X+\frac{4}{5}Y$ .

Substitute $\frac{4}{5}X-\frac{3}{5}Y$ , $\frac{3}{5}X+\frac{4}{5}Y$ for $x$ and $y$ respectively in the given equation.

  $\begin{array}{c}9x^2-24xy+16y^2=100\left(x-y-1\right)\\ 9{\left(\frac{4}{5}X-\frac{3}{5}Y\right)}^2-24\left(\frac{4}{5}X-\frac{3}{5}Y\right)\left(\frac{3}{5}X+\frac{4}{5}Y\right)+16{\left(\frac{3}{5}X+\frac{4}{5}Y\right)}^2=100\left(\left(\frac{4}{5}X-\frac{3}{5}Y\right)-\left(\frac{3}{5}X+\frac{4}{5}Y\right)-1\right)\\ \frac{9{\left(4X-3Y\right)}^2}{25}-\frac{24\left(4X-3Y\right)\left(3X+4Y\right)}{25}+\frac{16{\left(3X+4Y\right)}^2}{25}=100\left(\frac{1}{5}X-\frac{7}{5}Y-1\right)\\ \frac{9\left(16X^2+9Y^2-24XY\right)}{25}-\frac{24\left(12X^2+16XY-9XY-12Y^2\right)}{25}+\frac{16\left(9X^2+16Y^2+24XY\right)}{25}=100\left(\frac{X-7Y-5}{5}\right)\end{array}$

Further simplify,

  $\begin{array}{c}\frac{9\left(16X^2+9Y^2-24XY\right)}{25}-\frac{24\left(12X^2+16XY-9XY-12Y^2\right)}{25}+\frac{16\left(9X^2+16Y^2+24XY\right)}{25}=100\left(\frac{X-7Y-5}{5}\right)\\ 9\left(16X^2+9Y^2-24XY\right)-24\left(12X^2+16XY-9XY-12Y^2\right)+16\left(9X^2+16Y^2+24XY\right)-500\left(X-7Y-5\right)=0\\ X^2\left(144-288+144\right)+Y^2\left(81+288+256\right)+XY\left(-216-168+384\right)-500X+3500Y+2500=0\\ 625Y^2-500X+3500Y+2500=0\end{array}$

Further simplify,

  $\begin{array}{c}625Y^2-500X+3500Y+2500=0\\ {\left(5Y\right)}^2-20X+140Y+100=0\\ {\left(5Y\right)}^2+140Y+{\left(14\right)}^2-{\left(14\right)}^2=20X-100\\ {\left(5Y+14\right)}^2=20X+96\end{array}$

Therefore, the equation of the conic with the elimination of $xy$ term is equal to ${\left(5Y+14\right)}^2=20X+96$ .

(c)

To determine

To graph: The conic with equation $9x^2-24xy+16y^2=100\left(x-y-1\right)$ .

Explanation of Solution

Given information:

The equation of the conic is $9x^2-24xy+16y^2=100\left(x-y-1\right)$ .

Graph:

The graph of the conic with equation $9x^2-24xy+16y^2=100\left(x-y-1\right)$ is shown in figure below:

  

Figure(1)