Tangent PT and normal PN to the parabola `y^(2)=4ax` at a point P on it meet its axis at T and N respectively. The Locus of the centroid of the triangle PTN is a parabola whose

Options

A. vertex is `((2a)/(3),0)`
B. directrix is x = 0
C. latus rectum is `(2a)/(3)`
D. focus is (a, 0)

Answer

To find the locus of the centroid of triangle PTN formed by the tangent PT and normal PN to the parabola \( y^2 = 4ax \) at a point P on it, we can follow these steps:

### Step 1: Identify the point P on the parabola
Let the point \( P \) on the parabola \( y^2 = 4ax \) be represented in parametric form as:
\[ P(at^2, 2at) \]
where \( t \) is a parameter.

### Step 2: Find the equation of the tangent at point P
The equation of the tangent to the parabola at point \( P(at^2, 2at) \) is given by:
\[ y = tx + at \]

### Step 3: Find the coordinates of point T (intersection of tangent with x-axis)
To find point \( T \), set \( y = 0 \) in the tangent equation:
\[ 0 = tx + at \implies x = -\frac{at}{t} = -a \]
Thus, the coordinates of point \( T \) are:
\[ T(-a, 0) \]

### Step 4: Find the equation of the normal at point P
The equation of the normal to the parabola at point \( P(at^2, 2at) \) is given by:
\[ y - 2at = -\frac{1}{t}(x - at^2) \]
Rearranging gives:
\[ ty + 2at^2 = -x + at^2 \implies x + ty - at^2 - 2at = 0 \]

### Step 5: Find the coordinates of point N (intersection of normal with x-axis)
To find point \( N \), set \( y = 0 \) in the normal equation:
\[ x + 0 - at^2 - 2at = 0 \implies x = at^2 + 2at \]
Thus, the coordinates of point \( N \) are:
\[ N(at^2 + 2at, 0) \]

### Step 6: Find the centroid G of triangle PTN
The centroid \( G \) of triangle \( PTN \) is given by:
\[ G\left( \frac{x_P + x_T + x_N}{3}, \frac{y_P + y_T + y_N}{3} \right) \]
Substituting the coordinates:
\[ G\left( \frac{at^2 - a + (at^2 + 2at)}{3}, \frac{2at + 0 + 0}{3} \right) \]
This simplifies to:
\[ G\left( \frac{2at^2 + 2at - a}{3}, \frac{2at}{3} \right) \]

### Step 7: Express the coordinates of G in terms of t
Let’s express the coordinates of \( G \):
\[ G\left( \frac{2a(t^2 + t) - a}{3}, \frac{2at}{3} \right) \]

### Step 8: Eliminate t to find the locus
Let \( y = \frac{2at}{3} \) implies \( t = \frac{3y}{2a} \). Substitute this into the x-coordinate:
\[ x = \frac{2a\left(\left(\frac{3y}{2a}\right)^2 + \frac{3y}{2a}\right) - a}{3} \]
This simplifies to:
\[ x = \frac{2a\left(\frac{9y^2}{4a^2} + \frac{3y}{2a}\right) - a}{3} \]
\[ x = \frac{2a\left(\frac{9y^2 + 6ay}{4a^2}\right) - a}{3} \]
\[ x = \frac{18y^2 + 12ay - 4a^2}{12a} \]
This represents a parabola.

### Conclusion
The locus of the centroid \( G \) of triangle \( PTN \) is a parabola.