A trapezium is formed by the pair of tangents of parabola `P:y=(x^(2))/(4)+1` drawn from the centre of the ellipse `E:(x^(2))/(4)+y^(2)=(1)/(4)`, tangent at the vertex of P and the tangent at end point of the minor axis of E. The area (in sq. units) of trapezium is

Options

A. `(3)/(4)`
B. `(3)/(2)`
C. `(3)/(16)`
D. `(3)/(8)`

Answer

To find the area of the trapezium formed by the tangents to the parabola and the ellipse, we will follow these steps:

### Step 1: Identify the equations of the parabola and the ellipse.
The parabola is given by:
\[ P: y = \frac{x^2}{4} + 1 \]
The ellipse is given by:
\[ E: \frac{x^2}{4} + \frac{y^2}{1} = \frac{1}{4} \]

### Step 2: Determine 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 \(P\) can be found by setting \(x = 0\):
\[ y = \frac{0^2}{4} + 1 = 1 \]
Thus, the vertex of the parabola 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, and its endpoints can be found by setting \(x = 0\):
\[ \frac{0^2}{4} + \frac{y^2}{1} = \frac{1}{4} \implies y^2 = \frac{1}{4} \implies y = \pm \frac{1}{2} \]
Thus, the endpoints of the minor axis are \((0, \frac{1}{2})\) and \((0, -\frac{1}{2})\).

### Step 5: Find the equations of the tangents from the center of the ellipse to the parabola.
The general equation of the tangent to the parabola \(P\) at point \((x_1, y_1)\) is:
\[ y - y_1 = m(x - x_1) \]
Since we are drawing tangents from the origin \((0, 0)\), we can set \(y = mx\).

Substituting \(y = mx\) into the parabola's equation:
\[ mx = \frac{x^2}{4} + 1 \implies \frac{x^2}{4} - mx + 1 = 0 \]
For this quadratic to have real roots, the discriminant must be zero:
\[ (-m)^2 - 4 \cdot \frac{1}{4} \cdot 1 = 0 \implies m^2 - 1 = 0 \implies m = \pm 1 \]
Thus, the equations of the tangents are:
\[ y = x \quad \text{and} \quad y = -x \]

### Step 6: Find the equation of the tangent at the vertex of the parabola.
The tangent at the vertex \((0, 1)\) is a horizontal line:
\[ y = 1 \]

### Step 7: Find the equation of the tangent at the endpoint of the minor axis of the ellipse.
The endpoint of the minor axis is \((0, \frac{1}{2})\). The tangent at this point is also a horizontal line:
\[ y = \frac{1}{2} \]

### Step 8: Determine the area of the trapezium.
The trapezium is formed between the lines \(y = 1\) and \(y = \frac{1}{2}\) with the vertical distance between them being:
\[ \text{Height} = 1 - \frac{1}{2} = \frac{1}{2} \]
The lengths of the bases of the trapezium are:
- Base 1 (from \(y = 1\)): The points where the tangents intersect the line \(y = 1\) are \((1, 1)\) and \((-1, 1)\), so the length is \(2\).
- Base 2 (from \(y = \frac{1}{2}\)): The points where the tangents intersect the line \(y = \frac{1}{2}\) are \((\frac{1}{2}, \frac{1}{2})\) and \((- \frac{1}{2}, \frac{1}{2})\), so the length is \(1\).

The area \(A\) of the trapezium is given by:
\[ A = \frac{1}{2} \times (\text{Base 1} + \text{Base 2}) \times \text{Height} = \frac{1}{2} \times (2 + 1) \times \frac{1}{2} = \frac{3}{4} \text{ sq. units} \]

### Final Answer:
The area of the trapezium is \(\frac{3}{4}\) sq. units.