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The equation of the normal to the curve $y^{2} = 8x$ at the origin is ____________ .

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$y^{2} = 8x$

$\\2y \frac{dy}{dx} = 8\\ \frac{dy}{dx} = \frac{4}{y}\\$

$$Slope of normal at (0,0) $ = -\frac{y}{4} = 0 \\$

Equation of normal: y = 0

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