To solve the problem step by step, we will find the area of the trapezium formed by the tangents of the parabola and the ellipse as described in the question.
### Step 1: Identify the Parabola and Ellipse Equations
The given parabola is:
\[
P: y = \frac{x^2}{4} + 1
\]
The given ellipse is:
\[
E: \frac{x^2}{4} + y^2 = \frac{1}{4}
\]
### Step 2: Find the Center of the Ellipse
The center of the ellipse \(E\) is at the origin (0, 0).
### Step 3: Find the Vertex of the Parabola
The vertex of the parabola \(P\) is at the point (0, 1).
### Step 4: Find the Endpoints of the Minor Axis of the Ellipse
The minor axis of the ellipse is vertical. The endpoints can be found by setting \(x = 0\):
\[
\frac{0^2}{4} + y^2 = \frac{1}{4} \implies y^2 = \frac{1}{4} \implies y = \pm \frac{1}{2}
\]
Thus, the endpoints of the minor axis are (0, 1/2) and (0, -1/2).
### Step 5: Find the Tangent at the Vertex of the Parabola
The tangent at the vertex (0, 1) of the parabola is a horizontal line:
\[
y = 1
\]
### Step 6: Find the Tangent at the Endpoint of the Minor Axis of the Ellipse
The tangent at the endpoint (0, 1/2) of the minor axis is also a horizontal line:
\[
y = \frac{1}{2}
\]
### Step 7: Find the Pair of Tangents from the Center of the Ellipse to the Parabola
To find the tangents from the center (0, 0) to the parabola, we use the formula for the tangent line to a parabola \(y = ax^2 + bx + c\) from a point \((x_0, y_0)\):
\[
y = mx + c
\]
Substituting the point (0, 0) into the parabola equation:
\[
0 = a(0^2) + b(0) + c \implies c = 0
\]
The equation becomes:
\[
y = mx
\]
Substituting into the parabola equation:
\[
mx = \frac{x^2}{4} + 1 \implies x^2 - 4mx + 4 = 0
\]
For tangents, the discriminant must be zero:
\[
(4m)^2 - 4 \cdot 1 \cdot 4 = 0 \implies 16m^2 - 16 = 0 \implies m^2 = 1 \implies m = \pm 1
\]
Thus, the equations of the tangents are:
\[
y = x \quad \text{and} \quad y = -x
\]
### Step 8: Identify the Points of Intersection
The points where the tangents intersect the lines \(y = 1\) and \(y = \frac{1}{2}\):
1. For \(y = x\):
- At \(y = 1\): \(x = 1 \Rightarrow (1, 1)\)
- At \(y = \frac{1}{2}\): \(x = \frac{1}{2} \Rightarrow \left(\frac{1}{2}, \frac{1}{2}\right)\)
2. For \(y = -x\):
- At \(y = 1\): \(x = -1 \Rightarrow (-1, 1)\)
- At \(y = \frac{1}{2}\): \(x = -\frac{1}{2} \Rightarrow \left(-\frac{1}{2}, \frac{1}{2}\right)\)
### Step 9: Calculate the Area of the Trapezium
The trapezium has vertices at:
- \(A(1, 1)\)
- \(B(-1, 1)\)
- \(C\left(\frac{1}{2}, \frac{1}{2}\right)\)
- \(D\left(-\frac{1}{2}, \frac{1}{2}\right)\)
The lengths of the parallel sides are:
- Length of \(AB = 2\) (from \((1, 1)\) to \((-1, 1)\))
- Length of \(CD = 1\) (from \(\left(\frac{1}{2}, \frac{1}{2}\right)\) to \(\left(-\frac{1}{2}, \frac{1}{2}\right)\))
The height (distance between the parallel lines) is:
\[
h = 1 - \frac{1}{2} = \frac{1}{2}
\]
Using the area formula for a trapezium:
\[
\text{Area} = \frac{1}{2} \times (AB + CD) \times h = \frac{1}{2} \times (2 + 1) \times \frac{1}{2} = \frac{3}{4}
\]
### Final Answer
The area of the trapezium is:
\[
\boxed{\frac{3}{4}} \text{ square units.}
\]
