The tangents and normals are drawn at the extremites of the latusrectum of the parabola `y^2=4x`. The area of quadrilateral so formed is `lamda`sq units, the value of `lamda` is

Answer

To solve the problem, we need to find the area of the quadrilateral formed by the tangents and normals at the extremities of the latus rectum of the parabola \(y^2 = 4x\).

### Step-by-Step Solution:

1. **Identify the Latus Rectum Points**:
The latus rectum of the parabola \(y^2 = 4x\) is vertical and passes through the focus of the parabola. The focus is at \((1, 0)\) and the endpoints of the latus rectum are \((1, 2)\) and \((1, -2)\). Thus, the points are:
- \(A(1, 2)\)
- \(D(1, -2)\)

2. **Find the Tangent at Point A**:
The slope of the tangent at point \(A(1, 2)\) can be derived using the derivative of the parabola:
\[ \frac{dy}{dx} = \frac{4}{2y} = \frac{4}{4} = 1 \]
The equation of the tangent line at point \(A\) is:
\[ y - 2 = 1(x - 1) \implies y = x + 1 \]
The tangent line intersects the x-axis at point \(B\):
\[ 0 = x + 1 \implies x = -1 \implies B(-1, 0) \]

3. **Find the Normal at Point A**:
The slope of the normal is the negative reciprocal of the tangent slope:
\[ m = -1 \]
The equation of the normal line at point \(A\) is:
\[ y - 2 = -1(x - 1) \implies y = -x + 3 \]
The normal line intersects the x-axis at point \(C\):
\[ 0 = -x + 3 \implies x = 3 \implies C(3, 0) \]

4. **Identify the Points**:
We have the following points:
- \(A(1, 2)\)
- \(B(-1, 0)\)
- \(C(3, 0)\)
- \(D(1, -2)\)

5. **Calculate the Area of Quadrilateral ABCD**:
We can use the formula for the area of a quadrilateral given its vertices:
\[ \text{Area} = \frac{1}{2} \left| x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1) \right| \]
Substituting the coordinates:
\[ = \frac{1}{2} \left| 1\cdot0 + (-1)\cdot0 + 3\cdot(-2) + 1\cdot2 - (2\cdot(-1) + 0\cdot3 + 0\cdot1 + (-2)\cdot1) \right| \]
Simplifying:
\[ = \frac{1}{2} \left| 0 + 0 - 6 + 2 - (-2) \right| = \frac{1}{2} \left| -4 + 2 \right| = \frac{1}{2} \left| -4 \right| = \frac{1}{2} \cdot 4 = 8 \]

Thus, the area of the quadrilateral ABCD is \(8\) square units, so \(\lambda = 8\).

### Final Answer:
\[ \lambda = 8 \]