If $α$ is the inclination of a tangent to the parabola $y^{2} = 4 a x$ then the distance between the tangent and a parallel normal is

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Solution

The correct option is D
Let the tangent at $P (a t^{2} .2 a t)$ be parallel to the normal at $Q (t_{1}^{2}, 2 a t_{1})$
Slope of the tangent at P is $\frac{l}{t}$
Slope of the normal at Q is $- t_{1}$
Tangent at P is parallel to the normal at $Q \Rightarrow \frac{l}{t} = - t_{1} \Rightarrow t_{1} = - \frac{1}{t}$
$∴ Q = [a {(- \frac{l}{t})}^{2}, 2 a (- \frac{l}{t})] = (\frac{a}{t^{2}}, \frac{- 2}{t})$
Inclination of the tangent at P is $α \Rightarrow \frac{1}{t} = t a n α t = c o t α$ Distance between the tangent and parallel normal is
$P Q = \sqrt{{(a t^{2} - \frac{a}{t^{2}})}^{2} + {(2 a t^{2} - \frac{2 a}{t})}^{2}} = \sqrt{a^{2} {(t^{2} - \frac{l}{t^{2}})}^{2} + 4 a^{2} {(t + \frac{l a}{t})}^{2}}$
$= \sqrt{a^{2} {(t - \frac{l}{t^{2}})}^{2} {(t + \frac{l}{t})}^{2} + {(t + \frac{l}{t})}^{2}}$
$= \sqrt{a^{2} {(t + \frac{l}{t})}^{2} [{(t - \frac{l}{t})}^{2} + 4]} = \sqrt{a^{2} {(t + \frac{l}{t})}^{2} {(t + \frac{t + l}{t})}^{2}} = a (c o t α + t a n α)^{2} = a s e c^{2} α . C o s e c^{2} α$

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