If normal to parabola `y^(2)=4ax` at point `P(at^(2),2at)` intersects the parabola again at Q, such that sum of ordinates of the points P and Q is 3, then find the length of latus ectum in terms of t.

Explanation

To solve the problem step by step, we will analyze the given parabola and the properties of the normal line at a point on the parabola.

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

**Hint:** The coordinates of point \( P \) can be derived from the parametric equations of the parabola.

### Step 2: Write the equation of the normal at point \( P \)
The slope of the tangent to the parabola at point \( P \) is given by \( \frac{dy}{dx} = \frac{2a}{y} \). At point \( P(at^2, 2at) \), the slope of the tangent is \( \frac{2a}{2at} = \frac{1}{t} \). Therefore, the slope of the normal is the negative reciprocal, which is \( -t \).

Using the point-slope form of the equation of a line, the equation of the normal at point \( P \) is:
\[ y - 2at = -t(x - at^2) \]
Rearranging gives:
\[ y = -tx + at^3 + 2at \]

**Hint:** Remember that the normal line has a slope that is the negative reciprocal of the tangent slope.

### Step 3: Find the intersection of the normal with the parabola
To find the intersection point \( Q \) of the normal with the parabola, substitute \( y \) from the normal's equation into the parabola's equation:
\[ (-tx + at^3 + 2at)^2 = 4ax \]
Expanding and rearranging gives a quadratic equation in \( x \).

**Hint:** You might need to expand the left-hand side and collect like terms to form a standard quadratic equation.

### Step 4: Use the condition on the sum of ordinates
Let the coordinates of point \( Q \) be \( Q(at_1^2, 2at_1) \). According to the problem, the sum of the ordinates of points \( P \) and \( Q \) is given as:
\[ 2at + 2at_1 = 3 \]
This simplifies to:
\[ t + t_1 = \frac{3}{2a} \]

**Hint:** This equation relates the parameters \( t \) and \( t_1 \).

### Step 5: Solve for \( t_1 \)
From the equation \( t + t_1 = \frac{3}{2a} \), we can express \( t_1 \) in terms of \( t \):
\[ t_1 = \frac{3}{2a} - t \]

**Hint:** Substitute this expression back into your earlier equations to find relationships involving \( a \).

### Step 6: Find the length of the latus rectum
The length of the latus rectum of the parabola \( y^2 = 4ax \) is given by \( 4a \). We need to express \( a \) in terms of \( t \). From the earlier derived relationships, we can find that:
\[ 4a = -3t \]
Taking the modulus gives:
\[ \text{Length of latus rectum} = |4a| = |3t| \]

**Hint:** Remember that lengths are always positive, so take the absolute value.

### Final Answer
The length of the latus rectum in terms of \( t \) is:
\[ \text{Length of latus rectum} = 3|t| \]