From a point `(sintheta,costheta)`, if three normals can be drawn to the parabola `y^(2)=4ax` then the value of a is

Options

A. (1/2,1)
B. [-1/2,0)
C. [1/2,1]
D. `(-1/2,0cup(0,12))`

Explanation

To solve the problem of finding the value of \( a \) such that three normals can be drawn from the point \( (\sin \theta, \cos \theta) \) to the parabola \( y^2 = 4ax \), we will follow these steps:

### Step 1: Understand the Condition for Three Normals
For three normals to be drawn from a point \( (h, k) \) to the parabola \( y^2 = 4ax \), the x-coordinate \( h \) must satisfy the condition:
\[ h > 2a \]
In our case, \( h = \sin \theta \) and \( k = \cos \theta \).

### Step 2: Set Up the Inequality
We substitute \( h \) into the inequality:
\[ \sin \theta > 2a \]

### Step 3: Consider the Circle
The point \( (\sin \theta, \cos \theta) \) lies on the unit circle defined by:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
We need to find the intersection of the vertical line \( x = 2a \) with this circle.

### Step 4: Substitute into the Circle Equation
We substitute \( x = 2a \) into the equation of the circle:
\[ (2a)^2 + y^2 = 1 \]
This simplifies to:
\[ 4a^2 + y^2 = 1 \]
Rearranging gives:
\[ y^2 = 1 - 4a^2 \]

### Step 5: Condition for Two Intersection Points
For the line \( x = 2a \) to intersect the circle at two points, the right-hand side must be positive:
\[ 1 - 4a^2 > 0 \]
This leads to:
\[ 4a^2 < 1 \]

### Step 6: Solve the Inequality
Dividing both sides by 4 gives:
\[ a^2 < \frac{1}{4} \]
Taking the square root yields:
\[ |a| < \frac{1}{2} \]

### Step 7: Write the Final Result
This means:
\[ -\frac{1}{2} < a < \frac{1}{2} \]
However, since \( a \) must be positive for the parabola to be valid, we restrict our solution to:
\[ 0 < a < \frac{1}{2} \]

### Conclusion
Thus, the value of \( a \) such that three normals can be drawn from the point \( (\sin \theta, \cos \theta) \) to the parabola \( y^2 = 4ax \) is:
\[ a \in \left(0, \frac{1}{2}\right) \]