Locus of Centroid of Triangle Formed by Tangent and Normal to Parabola

To find the locus of the centroid of the triangle $PTN$, we first determine the coordinates of points $P$, $T$, and $N$.

Let the point $P$ on the parabola $y^2 = 4ax$ be parameterized as:
$P = (at^2, 2at)$

1. Coordinates of point $T$:
   The equation of the tangent to the parabola at $P$ is:
   $ty = x + at^2$
   The axis of the parabola is the $x$-axis ($y = 0$). Setting $y = 0$ gives:
   $x = -at^2$
   Thus, the point of intersection of the tangent with the axis is:
   $T = (-at^2, 0)$

2. Coordinates of point $N$:
   The equation of the normal to the parabola at $P$ is:
   $y + tx = 2at + at^3$
   Setting $y = 0$ for the intersection with the $x$-axis gives:
   $tx = 2at + at^3 \implies x = 2a + at^2$
   Thus, the point of intersection of the normal with the axis is:
   $N = (2a + at^2, 0)$

3. Centroid of $\triangle PTN$:
   Let the centroid of $\triangle PTN$ be $G(h, k)$. The coordinates of $G$ are:
   $h = \frac{x_P + x_T + x_N}{3} = \frac{at^2 + (-at^2) + (2a + at^2)}{3} = \frac{2a + at^2}{3}$
   $k = \frac{y_P + y_T + y_N}{3} = \frac{2at + 0 + 0}{3} = \frac{2at}{3}$

From the $y$-coordinate:
$t = \frac{3k}{2a}$

Substituting $t$ into the equation for $h$:
$3h = 2a + a\left(\frac{3k}{2a}\right)^2 = 2a + \frac{9k^2}{4a}$
$12ah = 8a^2 + 9k^2$
$9k^2 = 12a\left(h - \frac{2a}{3}\right)$

Replacing $(h, k)$ with $(x, y)$, the equation of the locus is:
$y^2 = \frac{4a}{3}\left(x - \frac{2a}{3}\right)$

This represents a parabola of the form $Y^2 = 4AX$, where:
$Y = y, \quad X = x - \frac{2a}{3}, \quad 4A = \frac{4a}{3} \implies A = \frac{a}{3}$

Now, analyzing the characteristics of this parabola:

• Vertex:
  $X = 0 \implies x = \frac{2a}{3}, \quad Y = 0 \implies y = 0$
  The vertex is $\left(\frac{2a}{3}, 0\right)$. (Option A is correct)

• Length of Latus Rectum:
  $\text{Latus Rectum} = 4A = \frac{4a}{3}$
  (Option C is incorrect)

• Directrix:
  $X = -A \implies x - \frac{2a}{3} = -\frac{a}{3} \implies x = \frac{a}{3}$
  (Option B is incorrect)

• Focus:
  $X = A \implies x - \frac{2a}{3} = \frac{a}{3} \implies x = a$
  $Y = 0 \implies y = 0$
  The focus is $(a, 0)$. (Option D is correct)

Therefore, the correct options are A and D.