The distance between a tangent to the parabola $y^{2} = 4 A x (A > 0)$ and the parallel normal with gradient $1$ is

4 A

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2 \sqrt{2} A

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2 A

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\sqrt{2} A

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Solution

The correct option is A $\sqrt{2} A$
The distance between the tangent to the parabola $y^{2} = 4 a x (A > 0)$
And the parallel normal with gradient $1$ is
Equation of tangent is
$t y = x + A t^{2}$
Slope $= \frac{1}{t} = 1 \Rightarrow t = 1$
Point of contact $\equiv (A t^{2}, 2 A t) = (A, 2 A)$
Equation of normal passes through focus $(A, 0)$ and gradient $1$ is
$y - 0 = 1 (x - A)$
$y = x - A$
Perpendicular distance at point $(A, 2 A)$ from normal $x - y - A = 0$ is
$\frac{| A - 2 A - A |}{\sqrt{1^{2} + 1^{2}}} = \frac{2 A}{\sqrt{2}} = \sqrt{2} A$

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