Chapter 11.5, Problem 36E

To determine

To show the equation $\sqrt{x}+\sqrt{y}=1$ is part of a parabola by rotate the axes through an angle of ${45}^0$ .

Explanation of Solution

Given information: Given equation $\sqrt{x}+\sqrt{y}=1$ .

Calculation:

From the given equation of parabola, first remove the radicals from the given equation by square both sides.

  $\begin{array}{l}{\left(\sqrt{x}+\sqrt{y}\right)}^2=1\\ x+y+2\sqrt{xy}=1\\ 2\sqrt{xy}=1-\left(x+y\right)\end{array}$

Since, this equation also contains radicals, again square the equation both sides.

  $\begin{array}{l}{\left(2\sqrt{xy}\right)}^2={\left(1-\left(x+y\right)\right)}^2\\ 4xy=1+{\left(x+y\right)}^2-2\left(x+y\right)\\ 4xy=1+x^2+y^2+2xy-2x-2y.\end{array}$

On shift all terms to the left side, to get

  $\begin{array}{l}4xy-1-x^2-y^2-2xy+2x+2y=0\\ -x^2-y^2+2xy+2x+2y-1=0\\ -\left(x^2-2xy+y^2-2x-2y+1\right)=0\\ x^2-2xy+y^2-2x-2y+1=0\end{array}$

The equation represents parabola, need to eliminates the $xy-\text{terms}$ by rotate the axes through an angle of ${45}^0$ .

  $\varphi ={45}^0$

On substitute $\varphi ={45}^0$ in the rotation of axes formulas given us,

On substitute the values of $X\text{ and }Y$ in the given equation, to get on combine the function,

  $\begin{array}{l}\frac{1}{2}\left[X^2+Y^2-2XY-2X^2+2Y^2+X^2+Y^2+2XY-2\sqrt{2}X+2\sqrt{2}Y-2\sqrt{2}X-2\sqrt{2}Y+2\right]=0\\ 4Y^2-4\sqrt{2}X+2=0\\ 2Y^2-2\sqrt{2}X=-1\\ Y^2=4\left(\frac{1}{2}\right)\left(\sqrt{2}X-\frac{1}{2}\right)\end{array}$

Since, this equation is same as the equation of standards from of the parabola as $y^2=4px$ ,

Thus, the equation represents a parabola.