A particle is moving in the xy - plane along a curve C passing through the point (3,3) .The tangent to the curve C at the point P meets the x - axis at Q . If the y - axis bisects the segment PQ, then C is parabola with

Options

A. length of latus rectum 3
B. length of latus rectum 6
C. focus `(4/3,0)`
D. focus `(0,3/3)`

Answer

To solve the problem, we need to determine the equation of the parabola \( C \) that passes through the point \( (3, 3) \) and satisfies the condition that the y-axis bisects the segment \( PQ \) where \( P \) is a point on the curve and \( Q \) is the point where the tangent at \( P \) meets the x-axis.

### Step-by-step Solution:

1. **Assume the form of the parabola**:
We can assume the parabola is of the form \( y^2 = 4ax \).

2. **Identify the point P**:
Let \( P \) be a point on the parabola. We can express \( P \) in terms of a parameter \( t \):
\[ P = (at^2, 2at) \]

3. **Find the equation of the tangent at point P**:
The equation of the tangent to the parabola \( y^2 = 4ax \) at point \( P(at^2, 2at) \) is given by:
\[ yt = x + at^2 \]

4. **Determine where the tangent meets the x-axis (point Q)**:
To find the coordinates of point \( Q \), we set \( y = 0 \) in the tangent equation:
\[ 0 = x + at^2 \implies x = -at^2 \]
Thus, the coordinates of point \( Q \) are:
\[ Q = (-at^2, 0) \]

5. **Find the midpoint of segment PQ**:
The midpoint \( M \) of segment \( PQ \) is given by:
\[ M = \left( \frac{at^2 + (-at^2)}{2}, \frac{2at + 0}{2} \right) = \left( 0, at \right) \]

6. **Condition for the y-axis to bisect PQ**:
For the y-axis to bisect the segment \( PQ \), the x-coordinate of the midpoint \( M \) must be 0, which we have already established.

7. **Use the given point (3, 3)**:
Since the parabola passes through the point \( (3, 3) \), we substitute \( (x, y) = (3, 3) \) into the parabola equation:
\[ 3^2 = 4a \cdot 3 \implies 9 = 12a \implies a = \frac{3}{4} \]

8. **Write the final equation of the parabola**:
Substituting \( a \) back into the equation of the parabola:
\[ y^2 = 4 \cdot \frac{3}{4} x \implies y^2 = 3x \]

Thus, the equation of the parabola \( C \) is:
\[ \boxed{y^2 = 3x} \]