Question:

Step 1: Identify point P and the focus. For the parabola $y^2 = 4ax$, the focus is $S(a, 0)$. The point $P(at^2, 2at)$ lies on the parabola. Step 2: Find the focal distance (distance SP). Using the distance formula: $$SP = \sqrt{(at^2 - a)^2 + (2at - 0)^2}$$ $$SP = \sqrt{a^2(t^2 - 1)^2 + 4a^2t^2}$$ $$SP = \sqrt{a^2(t^4 - 2t^2 + 1) + 4a^2t^2}$$ $$SP = \sqrt{a^2(t^4 + 2t^2 + 1)}$$ $$SP = \sqrt{a^2(t^2 + 1)^2}$$ $$SP = a(t^2 + 1)$$ Step 3: Find the equation of the normal at $P(at^2, 2at)$. The slope of the tangent at $P$ is $\frac{dy}{dx} = \frac{2a}{2y} = \frac{a}{y}$. At $P(at^2, 2at)$, the slope of the tangent is $\frac{a}{2at} = \frac{1}{t}$. Therefore, the slope of the normal is $-t$. The equation of the normal is: $$y - 2at = -t(x - at^2)$$ $$y = -tx + at^3 + 2at$$ $$tx + y - at(t^2 + 2) = 0$$ Step 4: Determine the center and radius of the circle. The circle has diameter SP. Its center $C$ is the midpoint of $S(a, 0)$ and $P(at^2, 2at)$: $$C = \left(\frac{a + at^2}{2}, \frac{0 + 2at}{2}\right) = \left(\frac{a(1 + t^2)}{2}, at\right)$$ The radius $r$ is half the focal distance: $$r = \frac{SP}{2} = \frac{a(t^2 + 1)}{2}$$ Step 5: Find the perpendicular distance from the center C to the normal line. The equation of the normal line is $tx + y - at(t^2 + 2) = 0$. The center of the circle is $C\left(\frac{a(1 + t^2)}{2}, at\right)$. The perpendicular distance $d$ from $C$ to the normal line is: $$d = \frac{\left|t \cdot \frac{a(1 + t^2)}{2} + at - at(t^2 + 2)\right|}{\sqrt{t^2 + 1^2}}$$ $$d = \frac{\left|\frac{at(1 + t^2)}{2} + at - at^3 - 2at\right|}{\sqrt{t^2 + 1}}$$ $$d = \frac{\left|\frac{at + at^3 + 2at - 2at^3 - 4at}{2}\right|}{\sqrt{t^2 + 1}}$$ $$d = \frac{\left|\frac{-at - at^3}{2}\right|}{\sqrt{t^2 + 1}}$$ $$d = \frac{\left|\frac{-at(1 + t^2)}{2}\right|}{\sqrt{t^2 + 1}}$$ Since $a > 0$ and $1 + t^2 > 0$: $$d = \frac{|at|(1 + t^2)}{2\sqrt{1 + t^2}}$$ $$d = \frac{|at|\sqrt{1 + t^2}}{2}$$ Step 6: Calculate the length of the intercept (chord length). The length of the intercept on the normal made by the circle is the length of the chord. Using the formula for chord length $2\sqrt{r^2 - d^2}$: $$r^2 = \left(\frac{a(t^2 + 1)}{2}\right)^2 = \frac{a^2(t^2 + 1)^2}{4}$$ $$d^2 = \left(\frac{|at|\sqrt{1 + t^2}}{2}\right)^2 = \frac{a^2t^2(1 + t^2)}{4}$$ Chord length $= 2\sqrt{\frac{a^2(t^2 + 1)^2}{4} - \frac{a^2t^2(1 + t^2)}{4}}$ $$= 2\sqrt{\frac{a^2(1 + t^2)}{4} [(t^2 + 1) - t^2]}$$ $$= 2\sqrt{\frac{a^2(1 + t^2)}{4} [1]}$$ $$= 2 \cdot \frac{a\sqrt{1 + t^2}}{2}$$ $$= a\sqrt{1 + t^2}$$