Chapter 10, Problem 10CT

To determine

The analysis of the equation $41x^2-24xy+34y^2-25=0$ by rotating the axis, so that there is no $xy-$ term. Graph the new equation.

Answer to Problem 10CT

Solution:

The equation is $\frac{x{\text{'}}^2}{1}+\frac{y{\text{'}}^2}{\frac{1}{2}}=1$ .

The graph of the equation $\frac{x{\text{'}}^2}{1}+\frac{y{\text{'}}^2}{\frac{1}{2}}=1$ is

  

Explanation of Solution

Given information:

The equation $41x^2-24xy+34y^2-25=0$

To transform the equation of the form $A{x}^2+Bxy+C{y}^2++Dx+Ey+F=0,B\ne 0$ into an equation in $x\text{'}$ and $y\text{'}$ without $x\text{'}y\text{'}-$ term, rotate the axis through an angle $\theta$ that satisfies the equation $\cot (2\theta )=\frac{A-C}{B}$ .

Compare the equation $41x^2-24xy+34y^2-25=0$ with $A{x}^2+Bxy+C{y}^2++Dx+Ey+F=0,B\ne 0$ gives $A=41,B=-24$ and $C=34$ ,

  $\Rightarrow \cot (2\theta )=\frac{41-34}{-24}$

  $\Rightarrow \cot (2\theta )=-\frac{7}{24}$

Since $\cot (2\theta )<0$, then ${90}^{\circ }<2\theta <{180}^{\circ }$

  $\Rightarrow {45}^{\circ }<\theta <{90}^{\circ }$ .

  

  $\cot (2\theta )=-\frac{7}{24}$ it follows that $\cos (2\theta )=-\frac{7}{25}$

  $\sin \theta =\sqrt{\frac{1-\cos (2\theta )}{2}}=\sqrt{\frac{1-\left(-\frac{7}{25}\right)}{2}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$

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

  $\theta ={\cos }^{-1}\left(\frac{3}{5}\right)\approx {53.13}^{°}$

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

  $\Rightarrow x=\frac{3}{5}x\text{'}-\frac{4}{5}y\text{'}$ and $y=\frac{4}{5}x\text{'}+\frac{3}{5}y\text{'}$

Substituting these values in the equation $41x^2-24xy+34y^2-25=0$ ,

  $\Rightarrow 41{\left(\frac{3}{5}x\text{'}-\frac{4}{5}y\text{'}\right)}^2-24\left(\frac{3}{5}x\text{'}-\frac{4}{5}y\text{'}\right)\left(\frac{4}{5}x\text{'}+\frac{3}{5}y\text{'}\right)+34{\left(\frac{4}{5}x\text{'}+\frac{3}{5}y\text{'}\right)}^2-25=0$

By simplifying,

  $\Rightarrow 41\frac{{\left(3x\text{'}-4y\text{'}\right)}^2}{25}-24\frac{\left(3x\text{'}-4y\text{'}\right)\left(4x\text{'}+3y\text{'}\right)}{25}+34\frac{{\left(4x\text{'}+3y\text{'}\right)}^2}{25}=25$

Multiplying by $25$ to both sides,

  $\Rightarrow 41{\left(3x\text{'}-4y\text{'}\right)}^2-24\left(3x\text{'}-4y\text{'}\right)\left(4x\text{'}+3y\text{'}\right)+34{\left(4x\text{'}+3y\text{'}\right)}^2=625$

  $\Rightarrow 41\left(9x{\text{'}}^2-24x\text{'}y\text{'}+16y{\text{'}}^2\right)-24\left(12x{\text{'}}^2-7x\text{'}y\text{'}-12y{\text{'}}^2\right)+34\left(16x{\text{'}}^2+24x\text{'}y\text{'}+9y{\text{'}}^2\right)=625$

  $\Rightarrow 369x{\text{'}}^2-574x\text{'}y\text{'}+656y{\text{'}}^2-288x{\text{'}}^2+169x\text{'}y\text{'}+288y{\text{'}}^2+544x{\text{'}}^2+816x\text{'}y\text{'}+306y{\text{'}}^2=625$

  $\Rightarrow 625x{\text{'}}^2+1250y{\text{'}}^2=625$

Dividing by $625$ on both sides of equation,

  $\Rightarrow x{\text{'}}^2+2y{\text{'}}^2=1$

  $\Rightarrow \frac{x{\text{'}}^2}{1}+\frac{y{\text{'}}^2}{\frac{1}{2}}=1$

This is an equation of an ellipse with center at $(h,k)=(0,0)$ in the $x\text{'}y\text{'}-$ plane.

Thus, $a=\sqrt{1}=1$, $b=\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}$ and the vertices are $(h+a,k)=(1,0)$, $(h-a,k)=(-1,0)$ in the $x\text{'}y\text{'}-$ plane.

  $c^2=a^2-b^2=1-\frac{1}{2}=\frac{1}{2}$

  $\Rightarrow c=\frac{\sqrt{2}}{2}$

The foci are at $(h\pm c,k)=\left(\pm \frac{\sqrt{2}}{2},0\right)$ in the $x\text{'}y\text{'}-$ plane.

The minor axis intercept are $(h,k\pm b)=\left(0,\pm \frac{\sqrt{2}}{2}\right)$

The graph of the equation $\frac{x{\text{'}}^2}{1}+\frac{y{\text{'}}^2}{\frac{1}{2}}=1$ is as follow: