Chapter 11.5, Problem 28E

(a)

To determine

To check: Whether the graph of the equation ${\left(7x+24y\right)}^2=600x-175y+25$ is a parabola, an ellipse or a hyperbola with the help of discriminant.

Answer to Problem 28E

The graph of the equation ${\left(7x+24y\right)}^2=600x-175y+25$ is a parabola.

Explanation of Solution

Given information:

The equation of the conic is ${\left(7x+24y\right)}^2=600x-175y+25$ .

Calculation:

Simplify the equation of the conic.

  $\begin{array}{c}{\left(7x+24y\right)}^2=600x-175y+25\\ 49x^2+576y^2+336xy=600x-175y+25\\ 49x^2+336xy+576y^2-600x+175y-25=0\end{array}$

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 $49$ , $336$ , $576$ , $-600$ , $175$ and $-25$ respectively.

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

Substitute $49$ , $336$ , $576$ for $A$ , $B$ , $C$ respectively in the discriminant.

  $\begin{array}{c}{B}^2-4AC={\left(336\right)}^2-4\times 49\times 576\\ =112896-112896\\ =0\end{array}$

The value of discriminant is equal to zero.

Therefore, the graph of the equation ${\left(7x+24y\right)}^2=600x-175y+25$ is a parabola.

(b)

To determine

To find: The equation of the conic ${\left(7x+24y\right)}^2=600x-175y+25$ with the elimination of $xy$ term with rotation of axes.

Answer to Problem 28E

The equation of the conic with the elimination of $xy$ term is equal to $336X+527Y=\sqrt{329375X-210000Y+15625}$ .

Explanation of Solution

Given information:

The equation of the conic is ${\left(7x+24y\right)}^2=600x-175y+25$ .

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 $49$ , $336$ , $576$ for $A$ , $B$ , $C$ in the equation for angle of rotation.

  $\begin{array}{c}\cot 2\varphi =\frac{A-C}{B}\\ =\frac{576-49}{336}\\ =\frac{527}{336}\end{array}$

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

Use the Pythagoras theorem for the hypotenuse.

  $\begin{array}{c}{H}^2=P^2+B^2\\ H^2={\left(527\right)}^2+{\left(336\right)}^2\\ H=\sqrt{277729+112896}\\ H=625\end{array}$

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 $\frac{527}{625}$ .

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.

  $\begin{array}{c}\cos \varphi =\sqrt{\frac{1+\cos 2\varphi }{2}}\\ =\sqrt{\frac{1+\frac{527}{625}}{2}}\\ =\sqrt{\frac{\frac{625+527}{625}}{2}}\\ =\sqrt{\frac{1152}{2\times 625}}\end{array}$

Further simplify,

  $\begin{array}{c}\cos \varphi =\sqrt{\frac{1152}{2\times 625}}\\ =\sqrt{\frac{576}{625}}\\ =\frac{24}{25}\end{array}$

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

  $\begin{array}{c}\sin \varphi =\sqrt{\frac{1-\cos 2\varphi }{2}}\\ =\sqrt{\frac{1-\frac{527}{625}}{2}}\\ =\sqrt{\frac{\frac{625-527}{625}}{2}}\\ =\sqrt{\frac{98}{2\times 625}}\end{array}$

Further simplify,

  $\begin{array}{c}\sin \varphi =\sqrt{\frac{98}{2\times 625}}\\ =\sqrt{\frac{49}{625}}\\ =\frac{7}{25}\end{array}$

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 $x=\frac{24}{25}X-\frac{7}{25}Y$ and $y=\frac{7}{25}X+\frac{24}{25}Y$ .

Substitute $\frac{24}{25}X-\frac{7}{25}Y$ , $\frac{7}{25}X+\frac{24}{25}Y$ for $x$ and $y$ respectively in the given equation.

  $\begin{array}{c}{\left(7\left(\frac{24}{25}X-\frac{7}{25}Y\right)+24\left(\frac{7}{25}X+\frac{24}{25}Y\right)\right)}^2=600\left(\frac{24}{25}X-\frac{7}{25}Y\right)-175\left(\frac{7}{25}X+\frac{24}{25}Y\right)+25\\ {\left(\frac{168}{25}X+\frac{168}{25}X-\frac{49}{25}Y+\frac{576}{25}Y\right)}^2=576X-168Y-49X-168Y+25\\ {\left(\frac{336}{25}X+\frac{527}{25}Y\right)}^2=527X-336Y+25\\ \frac{{\left(336X+527Y\right)}^2}{625}=527X-336Y+25\end{array}$

Further simplify,

  $\begin{array}{c}\frac{{\left(336X+527Y\right)}^2}{625}=527X-336Y+25\\ {\left(336X+527Y\right)}^2=329375X-210000Y+15625\\ 336X+527Y=\sqrt{329375X-210000Y+15625}\end{array}$

Therefore, the equation of the conic with the elimination of $xy$ term is equal to $336X+527Y=\sqrt{329375X-210000Y+15625}$ .

(c)

To determine

To graph: The conic with equation ${\left(7x+24y\right)}^2=600x-175y+25$ .

Explanation of Solution

Given information:

The equation of the conic is ${\left(7x+24y\right)}^2=600x-175y+25$ .

Graph:

The graph of the conic with equation ${\left(7x+24y\right)}^2=600x-175y+25$ is shown in figure below:

  

Figure(1)