Intersection Ordinate of Tangent and Normal to Parabola

To find the ordinate of the point of intersection between the tangent at point $P$ and the normal at point $S$, we first write down their respective equations for the parabola $y^2 = 4ax$.

Step 1: Equation of the tangent to the parabola at $P(at^2, 2at)$ The standard equation of the tangent at a parametric point $t$ is: $ty = x + at^2 \implies x = ty - at^2 \quad \text{--- (1)}$

Step 2: Equation of the normal to the parabola at $S(as^2, 2as)$ The standard equation of the normal at a parametric point $s$ is: $y + sx = 2as + as^3$

Given that $st = 1$, we have $s = \frac{1}{t}$. Substituting $s = \frac{1}{t}$ into the normal equation: $y + \frac{1}{t}x = \frac{2a}{t} + \frac{a}{t^3}$ Multiplying the entire equation by $t$: $ty + x = 2a + \frac{a}{t^2} \implies x = 2a + \frac{a}{t^2} - ty \quad \text{--- (2)}$

Step 3: Finding the ordinate ($y$-coordinate) of the intersection point Equating equation $(1)$ and equation $(2)$: $ty - at^2 = 2a + \frac{a}{t^2} - ty$

Rearranging the terms to solve for $y$: $2ty = at^2 + 2a + \frac{a}{t^2}$ $2ty = a\left(t^2 + 2 + \frac{1}{t^2}\right)$ $2ty = a\left(t + \frac{1}{t}\right)^2 = \frac{a(t^2 + 1)^2}{t^2}$

Dividing both sides by $2t$: $y = \frac{a(t^2 + 1)^2}{2t^3}$

Thus, the ordinate of the point of intersection is: $\frac{a(t^2+1)^2}{2t^3}$

Correct Answer: (B)