Chapter 8.4, Problem 39E

Solution

a)

To find: The type of the conic section for the equation $9x^2+24xy+16y^2+20x-15y+75=0$ .

The graph of equation $9x^2+24xy+16y^2+20x-15y+75=0$ is a parabola.

Given information:

The equation is $9x^2+24xy+16y^2+20x-15y+75=0$ .

Calculation:

Draw the graph of equation $9x^2+24xy+16y^2+20x-15y+75=0$

  

Therefore, the graph represents a parabola.

b)

To find: The equation in terms of $\left(u,v\right)$ with no cross-product term and using axis of rotation.

The required equation in terms of $\left(u,v\right)$ is $u^2=v-3$ .

Given information:

The equation is $9x^2+24xy+16y^2+20x-15y+75=0$ .

Formula used:

Formula used to find the rotation angle $\text{θ}$ is:

  $\cot \theta =\frac{A-C}{B}$ , $0<\theta <\frac{\pi }{2}$

Here, A , B , and C are the values obtain by comparing given equation with equation $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ .

The equations for the axis of rotation are,

  $x=u\cos \theta -v\sin \theta$

  $y=u\sin \theta +v\cos \theta$

Calculation:

Comparing the given equation with the general equation,

  $A=9$

  $B=24$

  $C=16$

Calculate the angle of rotation.

  $\begin{array}{c}\cot 2\theta =\frac{9-16}{24}\\ \cot 2\theta =\frac{-7}{24}\end{array}$

It is known that when $\cot 2\theta =\frac{-7}{24}$ , then $\cos 2\theta =-\frac{7}{25}$ .

Apply the trigonometric identity $\cos \theta =\sqrt{\frac{1+\cos 2\theta }{2}}$ and substitute $\cos 2\theta =-\frac{7}{25}$ into the identity and simplify.

  $\begin{array}{c}\cos \theta =\sqrt{\frac{1+\left(-\frac{7}{25}\right)}{2}}\\ =\sqrt{\frac{1-\frac{7}{25}}{2}}\\ =\sqrt{\frac{9}{25}}\\ =\frac{3}{5}\end{array}$

Similarly, apply the trigonometric identity $\sin \theta =\sqrt{\frac{1-\cos 2\theta }{2}}$ and substitute $\cos 2\theta =-\frac{7}{25}$ into the identity and simplify.

  $\begin{array}{c}\sin \theta =\sqrt{\frac{1-\left(-\frac{7}{25}\right)}{2}}\\ =\sqrt{\frac{1+\frac{7}{25}}{2}}\\ =\sqrt{\frac{16}{25}}\\ =\frac{4}{5}\end{array}$

Substitute $\sin \theta =\frac{\text{4}}{5}$ and $\cos \theta =\frac{\text{3}}{5}$ in above equation to find the equation in terms of $x$ .

  $\begin{array}{c}x=u\cdot \frac{3}{5}-v\cdot \frac{4}{5}\\ =\frac{3u}{5}-\frac{4v}{5}\end{array}$

Substitute $\sin \theta =\frac{\text{4}}{5}$ and $\cos \theta =\frac{\text{3}}{5}$ in above equation to find the equation in terms of $y$ .

  $\begin{array}{c}y=u\cdot \frac{4}{5}+v\cdot \frac{3}{5}\\ =\frac{4u}{5}+\frac{3v}{5}\end{array}$

Therefore, the required equations are $x=\frac{3u}{5}-\frac{4v}{5}$ and $y=\frac{4u}{5}+\frac{3v}{5}$ .

Substituting the values of x and y in given equation and simplify.

  $9{\left(\frac{3}{5}u-\frac{4}{5}v\right)}^2+24\left(\frac{3}{5}u-\frac{4}{5}v\right)\left(\frac{4}{5}u+\frac{3}{5}v\right)+16{\left(\frac{4}{5}u+\frac{3}{5}v\right)}^2+20\left(\frac{3}{5}u-\frac{4}{5}v\right)-15\left(\frac{4}{5}u+\frac{3}{5}v\right)-75=0$

On simplifying the above equation one will get $u^2=v-3$ .

The equation $9x^2+24xy+16y^2+20x-15y+75=0$ transfers in $\left(u,v\right)$ coordinates with no cross product as $u^2=v-3$ .

c)

To find: The vertex of equation $9x^2+24xy+16y^2+20x-15y+75=0$ in $\left(u,v\right)$ system.

The vertices of equation $9x^2+24xy+16y^2+20x-15y+75=0$ in $\left(u,v\right)$ system is $\left(\text{0,}3\right)$ .

Given information:

The equation is $9x^2+24xy+16y^2+20x-15y+75=0$ .

The equation in $\left(u,v\right)$ system is,

  $u^2=v-3$

Formula used:

The standard formula for aparabola is, ${\left(x-h\right)}^2=4p\left(y-k\right)$

Vertex of the parabolais: $\left(h,k\right)$

Calculation:

Comparing equation, $u^2=v-3$ with standard equation of parabola.

  $\begin{array}{c}h=0\\ k=3\\ p=\frac{1}{4}\end{array}$

Therefore, the vertex of the parabola is $\left(\text{0,}3\right)$ ..

d)

To find: The vertices of parabola in $\left(x,y\right)$ system.

The vertices of equation $9x^2+24xy+16y^2+20x-15y+75=0$ is $\left(-2.4,1.8\right)$ .

Given information:

The equation is $9x^2+24xy+16y^2+20x-15y+75=0$ .

The equations for the rotation of axis are, $x=\frac{3u}{5}-\frac{4v}{5}$ and $y=\frac{4u}{5}+\frac{3v}{5}$ .

The vertices of equation $9x^2+24xy+16y^2+20x-15y+75=0$ in $\left(u,v\right)$ system is $\left(\text{0,}3\right)$ .

Calculation:

Substitute the values $u=0$ and $v=3$ in the equation of rotation of axis.

  $\begin{array}{c}x=\frac{3}{5}\left(0\right)-\frac{4}{5}\left(3\right)\\ =\frac{-12}{5}\\ =-2.4\end{array}$

Similarly,

  $\begin{array}{c}y=\frac{4}{5}\left(0\right)+\frac{3}{5}\left(3\right)\\ =1.8\end{array}$

The vertices of equation $9x^2+24xy+16y^2+20x-15y+75=0$ is $\left(-2.4,1.8\right)$ .