The length of the Subnormal at point $P (t)$ on the parabola $y^{2} = 8 x$ is equal to

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Solution

Given, $y^{2} = 8 x \Rightarrow 2 y \frac{d y}{d x} = 8 \Rightarrow \frac{d y}{d x} = \frac{4}{y}$
$P (t) \equiv (a t^{2}, 2 a t) = (2 t^{2}, 4 t) (∵ 4 a = 8)$

$∴ m (slope of T) = \frac{4}{4 t} = \frac{1}{t}$

Length of subnormal $= | y_{1} \cdot m | = ∣ ∣ ∣ 4 t \times \frac{1}{t} ∣ ∣ ∣ = 4$ units.

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