Two tangents are drawn from a point `(-4, 3)` to the parabola `y^(2)=16x`. If `alpha` is the angle between them, then the value of `cos alpha` is

Options

A. 0
B. `(1)/(2)`
C. `(sqrt3)/(2)`
D. `(1)/(sqrt2)`

Answer

To solve the problem of finding the value of \( \cos \alpha \) where \( \alpha \) is the angle between the two tangents drawn from the point \((-4, 3)\) to the parabola \( y^2 = 16x \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Parabola**:
The given parabola is \( y^2 = 16x \). This can be rewritten in the standard form \( y^2 = 4ax \) where \( a = 4 \). The vertex of the parabola is at the point \( (0, 0) \).

2. **Equation of the Tangent**:
The equation of the tangent to the parabola \( y^2 = 4ax \) at a point \( (at^2, 2at) \) is given by:
\[ y = mx + \frac{a}{m} \]
where \( m \) is the slope of the tangent.

3. **Finding the Slope**:
Since the tangents pass through the point \((-4, 3)\), we substitute this point into the tangent equation:
\[ 3 = m(-4) + \frac{4}{m} \]
Rearranging gives:
\[ 3 = -4m + \frac{4}{m} \]
Multiplying through by \( m \) to eliminate the fraction:
\[ 3m = -4m^2 + 4 \]
Rearranging this into standard quadratic form:
\[ 4m^2 + 3m - 4 = 0 \]

4. **Solving the Quadratic Equation**:
We can use the quadratic formula \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 4, b = 3, c = -4 \):
\[ m = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 4 \cdot (-4)}}{2 \cdot 4} \]
\[ m = \frac{-3 \pm \sqrt{9 + 64}}{8} \]
\[ m = \frac{-3 \pm \sqrt{73}}{8} \]
Let \( m_1 = \frac{-3 + \sqrt{73}}{8} \) and \( m_2 = \frac{-3 - \sqrt{73}}{8} \).

5. **Finding the Angle Between the Tangents**:
The angle \( \alpha \) between the two tangents can be found using the formula:
\[ \tan \alpha = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
First, calculate \( m_1 m_2 \):
\[ m_1 m_2 = \frac{(-3 + \sqrt{73})(-3 - \sqrt{73})}{64} = \frac{9 - 73}{64} = \frac{-64}{64} = -1 \]
Now calculate \( m_1 - m_2 \):
\[ m_1 - m_2 = \frac{-3 + \sqrt{73}}{8} - \frac{-3 - \sqrt{73}}{8} = \frac{2\sqrt{73}}{8} = \frac{\sqrt{73}}{4} \]
Therefore,
\[ \tan \alpha = \left| \frac{\frac{\sqrt{73}}{4}}{1 - 1} \right| \text{ (undefined, as denominator is 0)} \]
This indicates that \( \alpha = 90^\circ \).

6. **Finding \( \cos \alpha \)**:
Since \( \alpha = 90^\circ \):
\[ \cos \alpha = \cos 90^\circ = 0 \]

### Final Answer:
\[ \cos \alpha = 0 \]