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 Equation of the Parabola
The given equation of the parabola is:
\[ y^2 - 4y - 4x + 8 = 0 \]
We can rearrange it to a standard form. Completing the square for the \( y \) terms:
\[ y^2 - 4y = (y - 2)^2 - 4 \]
Thus, the equation becomes:
\[ (y - 2)^2 - 4 - 4x + 8 = 0 \implies (y - 2)^2 = 4x - 4 \]
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: Find the Equation of the Tangents
The general equation of the tangent to the parabola \( y^2 = 4ax \) at a point \( (x_1, y_1) \) is given by:
\[ y - y_1 = m(x - x_1) \]
For our parabola, we have:
\[ y - 2 = m(x - 1) + \frac{1}{m} \]
Substituting the point (2, 6):
\[ 6 - 2 = m(2 - 1) + \frac{1}{m} \]
This simplifies to:
\[ 4 = m + \frac{1}{m} \]
Multiplying through by \( m \) gives:
\[ 4m = m^2 + 1 \implies m^2 - 4m + 1 = 0 \]

### Step 3: Solve for the Slopes of the Tangents
Using the quadratic formula to find \( m \):
\[ m = \frac{4 \pm \sqrt{(4)^2 - 4 \cdot 1}}{2} = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3} \]
Thus, the slopes of the tangents are:
\[ m_1 = 2 + \sqrt{3}, \quad m_2 = 2 - \sqrt{3} \]

### Step 4: Find the Angle Between the Tangents
The angle \( \theta \) between the two tangents can be found using the formula:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Calculating \( m_1 - m_2 \):
\[ m_1 - m_2 = (2 + \sqrt{3}) - (2 - \sqrt{3}) = 2\sqrt{3} \]
Calculating \( m_1 m_2 \):
\[ m_1 m_2 = (2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1 \]
Now substituting into the formula:
\[ \tan \theta = \left| \frac{2\sqrt{3}}{1 + 1} \right| = \left| \frac{2\sqrt{3}}{2} \right| = \sqrt{3} \]
Thus, the angle \( \theta \) is:
\[ \theta = 60^\circ \]

### Final Answer
The angle between the tangents drawn from the point (2, 6) to the parabola is \( 60^\circ \).