The angle between the tangents drawn from the point (2, 6) to the parabola `y^(2)-4y-4x+8=0` is

Options

A. `(pi)/(2)`
B. `(pi)/(4)`
C. `(pi)/(3)`
D. `(pi)/(6)`

Answer

To find the angle between the tangents drawn from the point (2, 6) to the parabola given by the equation \(y^2 - 4y - 4x + 8 = 0\), we can follow these steps:

### Step 1: Rewrite the parabola in standard form
We start with the equation of the parabola:
\[ y^2 - 4y - 4x + 8 = 0 \]
Rearranging gives:
\[ y^2 - 4y = 4x - 8 \]
Completing the square for the left side:
\[ (y - 2)^2 - 4 = 4x - 8 \]
Adding 4 to both sides:
\[ (y - 2)^2 = 4x - 4 \]
This can be rewritten as:
\[ (y - 2)^2 = 4(x - 1) \]
This shows that the parabola opens to the right with vertex at (1, 2).

### Step 2: Use the tangent equation
The general equation of the tangent to the parabola \(y^2 = 4ax\) is given by:
\[ y = mx + \frac{a}{m} \]
For our parabola, \(a = 1\) (since \(4a = 4\)), so the equation becomes:
\[ y = mx + \frac{1}{m} \]

### Step 3: Substitute the point (2, 6)
Since the tangents pass through the point (2, 6), we substitute \(x = 2\) and \(y = 6\) into the tangent equation:
\[ 6 = m(2) + \frac{1}{m} \]
This simplifies to:
\[ 6 = 2m + \frac{1}{m} \]
Multiplying through by \(m\) to eliminate the fraction:
\[ 6m = 2m^2 + 1 \]
Rearranging gives:
\[ 2m^2 - 6m + 1 = 0 \]

### Step 4: Solve the quadratic equation
Using the quadratic formula \(m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
Here, \(a = 2\), \(b = -6\), and \(c = 1\):
\[ m = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} \]
Calculating the discriminant:
\[ m = \frac{6 \pm \sqrt{36 - 8}}{4} = \frac{6 \pm \sqrt{28}}{4} = \frac{6 \pm 2\sqrt{7}}{4} = \frac{3 \pm \sqrt{7}}{2} \]
Thus, the slopes of the tangents are:
\[ m_1 = \frac{3 + \sqrt{7}}{2}, \quad m_2 = \frac{3 - \sqrt{7}}{2} \]

### Step 5: Find the angle between the tangents
The angle \(\theta\) between the two tangents can be found using the formula:
\[ \tan \theta = \frac{m_1 - m_2}{1 + m_1 m_2} \]
Calculating \(m_1 - m_2\):
\[ m_1 - m_2 = \left(\frac{3 + \sqrt{7}}{2}\right) - \left(\frac{3 - \sqrt{7}}{2}\right) = \frac{2\sqrt{7}}{2} = \sqrt{7} \]
Calculating \(m_1 m_2\):
\[ m_1 m_2 = \left(\frac{3 + \sqrt{7}}{2}\right) \left(\frac{3 - \sqrt{7}}{2}\right) = \frac{(3)^2 - (\sqrt{7})^2}{4} = \frac{9 - 7}{4} = \frac{2}{4} = \frac{1}{2} \]
Now substituting these into the tangent formula:
\[ \tan \theta = \frac{\sqrt{7}}{1 + \frac{1}{2}} = \frac{\sqrt{7}}{\frac{3}{2}} = \frac{2\sqrt{7}}{3} \]

### Step 6: Find \(\theta\)
To find \(\theta\), we can use the inverse tangent function:
\[ \theta = \tan^{-1}\left(\frac{2\sqrt{7}}{3}\right) \]

### Conclusion
The angle between the tangents drawn from the point (2, 6) to the parabola is:
\[ \theta = \tan^{-1}\left(\frac{2\sqrt{7}}{3}\right) \]