Chapter 11.5, Problem 14E

To determine

To find : the equation of the given conic in $XY$ - coordinates

Answer to Problem 14E

The equation of conics is $\left(\frac{x^2+y^2}{\sqrt{2}}\right)=0$

Explanation of Solution

Given information : When the coordinate axes are rotated through the indicated angle at $\varphi =\frac{\pi }{4}$ of equation $xy=x+y$

Calculation :

The equation of conic is obtained using the standard equation of coordinate of $\left(X,Y\right)$ ,

  $\begin{array}{l}xy=x+y\\ \varphi =\frac{\pi }{4}\Rightarrow {45}^{\circ }\\ X=x\cos \varphi -y\sin \varphi \\ Y=x\sin \varphi +y\cos \varphi \\ X=x\cos {45}^{\circ }-y\sin {45}^{\circ }\\ X=x\left(\frac{1}{\sqrt{2}}\right)-y\left(\frac{1}{\sqrt{2}}\right)\Rightarrow \frac{x-y}{\sqrt{2}}\\ Y=x\sin {45}^{\circ }+y\cos {45}^{\circ }\\ Y=x\left(\frac{1}{\sqrt{2}}\right)+y\left(\frac{1}{\sqrt{2}}\right)\Rightarrow \frac{x+y}{\sqrt{2}}\\ xy=x+y\end{array}$

Substitute the value of x and y ,

  $\begin{array}{l}\left(\frac{x-y}{\sqrt{2}}\right)\left(\frac{x+y}{\sqrt{2}}\right)=\left(\frac{x-y}{\sqrt{2}}\right)+\left(\frac{x+y}{\sqrt{2}}\right)\\ \left(\frac{x^2+xy-xy+y^2}{\sqrt{2}}\right)=\left(\frac{x-y+x+y}{\sqrt{2}}\right)\\ \Rightarrow \boxed{\left(\frac{x^2+y^2}{\sqrt{2}}\right)=0}\end{array}$

Hence,

The equation of conics is $\left(\frac{x^2+y^2}{\sqrt{2}}\right)=0$