A line bisecting the ordinate PN of a point `P(at^2,2at),t gt 0` , on the parabola `y^2=4ax` is drawn parallel to the axis to meet the curve at Q. If NQ meets the tangent at the vertex at the point T, then the coordinates of T are:

Options

A. `(0,(4/3)at)`
B. `(0,2at)`
C. `((1/4)at^2,at)`
D. `(0,at)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the coordinates of point P
Given point \( P(at^2, 2at) \) where \( t > 0 \), we can substitute the values:
- \( P = (at^2, 2at) \)

### Step 2: Find the coordinates of point N
Point N is the foot of the perpendicular from point P to the x-axis. Since the x-coordinate remains the same and the y-coordinate is 0, we have:
- \( N = (at^2, 0) \)

### Step 3: Determine the midpoint of PN
The ordinate of point P is \( 2at \). The line bisecting the ordinate PN will be at half the distance of the y-coordinate of P:
- Midpoint of PN = \( \left(at^2, \frac{2at}{2}\right) = (at^2, at) \)

### Step 4: Find the coordinates of point Q
Since the line bisecting PN is parallel to the y-axis, point Q will have the same x-coordinate as P and the y-coordinate of the midpoint:
- \( Q = (at^2, at) \)

### Step 5: Find the equation of line NQ
The coordinates of N are \( (at^2, 0) \) and Q are \( (at^2, at) \). The slope of line NQ is:
- Slope \( m = \frac{at - 0}{at^2 - at^2} = \text{undefined} \)
This means line NQ is vertical, and its equation is:
- \( x = at^2 \)

### Step 6: Find the equation of the tangent at the vertex
The vertex of the parabola \( y^2 = 4ax \) is at the origin (0,0). The equation of the tangent at the vertex is:
- \( y = 0 \) (the x-axis)

### Step 7: Find the intersection point T of line NQ and the tangent
Since line NQ is vertical at \( x = at^2 \) and the tangent is the x-axis (y = 0), we can find the coordinates of point T:
- \( T = (at^2, 0) \)

### Final Answer
The coordinates of point T are:
- \( T(at^2, 0) \)

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