From a point on the axis of x common tangents are drawn to the parabola the ellipse `y^(2)`=4x and the ellipse `x^(2)/a^(2)+y^(2)/b^(2)=1(agtbgt0)`. If these tangents from an equilateral trianlge with their chord of contact w.r.t parabola, then set of exhaustive values of a is

Options

A. `(0,3)`
B. `(0,3/2)`
C. `(3/2,3)`
D. `(1,3/2)`

Answer

To solve the problem step by step, we will analyze the given information and apply the necessary mathematical concepts.

### Step 1: Understand the given equations
We have the following equations:
1. The parabola: \( y^2 = 4x \)
2. The ellipse: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)

Where \( a > b > 0 \).

### Step 2: Identify the point from which tangents are drawn
Let the point from which the tangents are drawn be \( P(h, 0) \) on the x-axis.

### Step 3: Find the tangents to the parabola
The equation of the tangent to the parabola \( y^2 = 4x \) at point \( P(h, 0) \) can be expressed as:
\[ y = mx + \frac{1}{m} \]
where \( m \) is the slope of the tangent.

### Step 4: Determine the slope \( m \)
Since the tangents form an equilateral triangle with the x-axis, the angles at point \( P \) are \( 30^\circ \) each (since the total angle is \( 60^\circ \)). Thus, the slope \( m \) can be calculated as:
\[ m = \tan(30^\circ) = \frac{1}{\sqrt{3}} \]
The tangents can also have a negative slope:
\[ m = -\frac{1}{\sqrt{3}} \]

### Step 5: Use the tangency condition for the ellipse
For the ellipse, the tangency condition is given by:
\[ c^2 = a^2 m^2 + b^2 \]
Where \( c \) is the distance from the point \( P(h, 0) \) to the center of the ellipse, which is \( h \).

### Step 6: Substitute values into the tangency condition
Substituting \( c = h \) and \( m = \frac{1}{\sqrt{3}} \):
\[ h^2 = a^2 \left(\frac{1}{\sqrt{3}}\right)^2 + b^2 \]
\[ h^2 = \frac{a^2}{3} + b^2 \]

### Step 7: Express \( b^2 \) in terms of \( a \)
Using the relationship \( b^2 = a^2(1 - e^2) \) where \( e \) is the eccentricity of the ellipse:
\[ h^2 = \frac{a^2}{3} + a^2(1 - e^2) \]
This simplifies to:
\[ h^2 = \frac{a^2}{3} + a^2 - a^2 e^2 \]
\[ h^2 = a^2 \left(1 + \frac{1}{3} - e^2\right) \]
\[ h^2 = a^2 \left(\frac{4}{3} - e^2\right) \]

### Step 8: Set conditions for eccentricity
Since \( e^2 \) must be between 0 and 1:
\[ 0 < \frac{4}{3} - e^2 < 1 \]
This leads to:
1. \( e^2 < \frac{4}{3} \)
2. \( e^2 > \frac{1}{3} \)

### Step 9: Solve for \( a \)
From the conditions on \( e^2 \):
1. \( 0 < 4a^2 - 9 < 3a^2 \)
2. Rearranging gives:
- \( 4a^2 - 9 > 0 \) implies \( a^2 > \frac{9}{4} \) or \( a > \frac{3}{2} \)
- \( 4a^2 - 9 < 3a^2 \) implies \( a^2 < 9 \) or \( a < 3 \)

### Final Result
Thus, the exhaustive values of \( a \) are:
\[ \frac{3}{2} < a < 3 \]