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: (a) `(0,(4/3)at)` (b) `(0,2at)` (c) `((1/4)at^2,at)` (d) `(0,at)`

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:

1. **Identify the coordinates of point P**: The point \( P \) on the parabola \( y^2 = 4ax \) is given as \( P(at^2, 2at) \).

2. **Find the coordinates of point N**: The ordinate \( PN \) is a vertical line from point \( P \) down to the x-axis. The coordinates of point \( N \) will be \( N(at^2, 0) \).

3. **Determine the midpoint of PN**: The midpoint \( M \) of the line segment \( PN \) can be calculated as:
\[ M = \left( \frac{at^2 + at^2}{2}, \frac{2at + 0}{2} \right) = \left( at^2, at \right) \]

4. **Draw a line parallel to the x-axis through M**: The line parallel to the x-axis at \( y = at \) will intersect the parabola \( y^2 = 4ax \). To find the points of intersection, we set \( y = at \) in the parabola's equation:
\[ (at)^2 = 4ax \]
Simplifying this gives:
\[ a^2t^2 = 4ax \implies x = \frac{at^2}{4} \]
Therefore, the coordinates of point \( Q \) are \( Q\left(\frac{at^2}{4}, at\right) \).

5. **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 simply the x-axis, which is \( y = 0 \).

6. **Determine the coordinates of point T**: The line segment \( NQ \) connects points \( N(at^2, 0) \) and \( Q\left(\frac{at^2}{4}, at\right) \). The slope of line \( NQ \) is:
\[ \text{slope} = \frac{at - 0}{\frac{at^2}{4} - at^2} = \frac{at}{\frac{at^2}{4} - \frac{4at^2}{4}} = \frac{at}{-\frac{3at^2}{4}} = -\frac{4}{3t} \]
The equation of line \( NQ \) can be written in point-slope form:
\[ y - 0 = -\frac{4}{3t}(x - at^2) \]
Simplifying gives:
\[ y = -\frac{4}{3t}x + \frac{4}{3}at \]

7. **Find the intersection of line NQ with the tangent at the vertex**: Setting \( y = 0 \) in the equation of line \( NQ \):
\[ 0 = -\frac{4}{3t}x + \frac{4}{3}at \]
Solving for \( x \):
\[ \frac{4}{3t}x = \frac{4}{3}at \implies x = at^2 \]
Since \( y = 0 \), the coordinates of point \( T \) are \( (0, \frac{4}{3}at) \).

8. **Final answer**: The coordinates of point \( T \) are \( (0, \frac{4}{3}at) \).