Question:

Step 1: Determine the equation of the tangent with gradient 1. The equation of a tangent to the parabola $y^2 = 4Ax$ with gradient $m$ is given by $y = mx + \frac{A}{m}$. Given that the gradient of the tangent is $m=1$, the equation of the tangent is: $$y = (1)x + \frac{A}{1}$$ $$y = x + A$$ This can be written in the general form as $x - y + A = 0$. Step 2: Determine the equation of the normal with gradient 1. The problem specifies a normal that is parallel to the tangent, meaning the normal also has a gradient of 1. For a normal to have a gradient of 1, the tangent at the point of normalcy must have a gradient of $-1$ (since the product of gradients of perpendicular lines is $-1$). The gradient of the tangent to the parabola $y^2 = 4Ax$ at a point $(At^2, 2At)$ is $\frac{1}{t}$. Setting the tangent's gradient to $-1$: $$\frac{1}{t} = -1 \implies t = -1$$ The point on the parabola where this normal is drawn is $(A(-1)^2, 2A(-1)) = (A, -2A)$. Now, we write the equation of the normal passing through $(A, -2A)$ with a gradient of 1: $$y - (-2A) = 1(x - A)$$ $$y + 2A = x - A$$ $$y = x - 3A$$ This can be written in the general form as $x - y - 3A = 0$. Step 3: Calculate the distance between the parallel lines. We have two parallel lines: Tangent: $x - y + A = 0$ Normal: $x - y - 3A = 0$ The distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by the formula $\frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$. In this case, $A=1$, $B=-1$, $C_1=A$, and $C_2=-3A$. The distance $D$ is: $$D = \frac{|A - (-3A)|}{\sqrt{1^2 + (-1)^2}}$$ $$D = \frac{|A + 3A|}{\sqrt{1 + 1}}$$ $$D = \frac{|4A|}{\sqrt{2}}$$ Since $A > 0$, $|4A| = 4A$. $$D = \frac{4A}{\sqrt{2}} = \frac{4A\sqrt{2}}{2} = 2\sqrt{2}A$$