# Question #608e5

##### 1 Answer
Mar 1, 2017

${y}_{x} = \pm \frac{\sqrt{1 - 2 k x}}{k}$

#### Explanation:

$k = \frac{1}{x + \sqrt{{x}^{2} + {y}^{2}}}$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow x + \sqrt{{x}^{2} + {y}^{2}} = \frac{1}{k}$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow \sqrt{{x}^{2} + {y}^{2}} = \frac{1}{k} - x$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow {x}^{2} + {y}^{2} = \frac{1}{k} ^ 2 - 2 \frac{x}{k} + {x}^{2}$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow {y}^{2} = \frac{1}{k} ^ 2 - \frac{2 k x}{k} ^ 2$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow y = \pm \frac{\sqrt{1 - 2 k x}}{k}$

Assuming $k$ is a (non-zero) constant
and $y$ is dependent on the variable $x$
$\textcolor{w h i t e}{\text{XXX}} {y}_{x} = \pm \frac{\sqrt{1 - 2 k x}}{k}$