If sqrt{y + x} + sqrt{y - x} = c and frac{dy} | vedclass

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(D) Given $\sqrt{y + x} + \sqrt{y - x} = c$. Squaring both sides:$(y + x) + (y - x) + 2\sqrt{(y + x)(y - x)} = c^2$$2y + 2\sqrt{y^2 - x^2} = c^2$$\sqr

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$\frac{y}{x}$

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(D) Given $\sqrt{y + x} + \sqrt{y - x} = c$. Squaring both sides:$(y + x) + (y - x) + 2\sqrt{(y + x)(y - x)} = c^2$$2y + 2\sqrt{y^2 - x^2} = c^2$$\sqrt{y^2 - x^2} = \frac{c^2 - 2y}{2} = \frac{c^2}{2} - y$Squaring again:$y^2 - x^2 = (\frac{c^2}{2})^2 - c^2y + y^2$$-x^2 = \frac{c^4}{4} - c^2y$$c^2y = x^2 + \frac{c^4}{4}$Differentiating with respect to $x$:$c^2 \frac{dy}{dx} = 2x$$\frac{dy}{dx} = \frac{2x}{c^2}$From $\sqrt{y^2 - x^2} = \frac{c^2}{2} - y$, we have $c^2 = 2(\sqrt{y^2 - x^2} + y)$.Substitute $c^2$ into $\frac{dy}{dx}$:$\frac{dy}{dx} = \frac{2x}{2(\sqrt{y^2 - x^2} + y)} = \frac{x}{\sqrt{y^2 - x^2} + y} = \frac{x/y}{\sqrt{(y^2 - x^2)/y^2} + 1} = \frac{x/y}{1 + \sqrt{1 - (x/y)^2}}$This does not match the form directly. Let's re-evaluate $\frac{dy}{dx}$ from $y^2 - x^2 = (\frac{c^2}{2} - y)^2$:$2y \frac{dy}{dx} - 2x = 2(\frac{c^2}{2} - y)(-\frac{dy}{dx})$$y \frac{dy}{dx} - x = -(\frac{c^2}{2} - y) \frac{dy}{dx} = -\sqrt{y^2 - x^2} \frac{dy}{dx}$$\frac{dy}{dx} (y + \sqrt{y^2 - x^2}) = x$$\frac{dy}{dx} = \frac{x}{y + \sqrt{y^2 - x^2}} = \frac{x}{y + y\sqrt{1 - (x/y)^2}} = \frac{x/y}{1 + \sqrt{1 - (x/y)^2}}$.Actually, the expression $\frac{dy}{dx} = \frac{x}{y + \sqrt{y^2 - x^2}}$ can be rationalized as $\frac{x(y - \sqrt{y^2 - x^2})}{y^2 - (y^2 - x^2)} = \frac{x(y - \sqrt{y^2 - x^2})}{x^2} = \frac{y - \sqrt{y^2 - x^2}}{x} = \frac{y}{x} - \sqrt{\frac{y^2}{x^2} - 1}$.Comparing with $\frac{dy}{dx} = K - \sqrt{\frac{y^2}{x^2} - 1}$, we get $K = \frac{y}{x}$.

If $\sqrt{y + x} + \sqrt{y - x} = c$ and $\frac{dy}{dx} = K - \sqrt{\frac{y^2}{x^2} - 1}$, then the value of $K$ is

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