The angle between the normals to the parabola `y^(2)=24x` at points (6, 12) and (6, -12), is

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

To find the angle between the normals to the parabola \( y^2 = 24x \) at the points \( (6, 12) \) and \( (6, -12) \), we will follow these steps:

### Step 1: Find the derivative of the parabola
The equation of the parabola is given by \( y^2 = 24x \). To find the slope of the tangent at any point on the parabola, we differentiate it with respect to \( x \).

\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(24x) \]

Using implicit differentiation:

\[ 2y \frac{dy}{dx} = 24 \]

Thus, we can solve for \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{24}{2y} = \frac{12}{y} \]

### Step 2: Calculate the slope of the tangent at the point \( (6, 12) \)
Substituting \( y = 12 \) into the derivative:

\[ \frac{dy}{dx} \bigg|_{(6, 12)} = \frac{12}{12} = 1 \]

### Step 3: Find the slope of the normal at the point \( (6, 12) \)
The slope of the normal is the negative reciprocal of the slope of the tangent. If the slope of the tangent \( m_1 = 1 \), then the slope of the normal \( m_2 \) is:

\[ m_2 = -\frac{1}{m_1} = -1 \]

### Step 4: Calculate the slope of the tangent at the point \( (6, -12) \)
Now, substituting \( y = -12 \) into the derivative:

\[ \frac{dy}{dx} \bigg|_{(6, -12)} = \frac{12}{-12} = -1 \]

### Step 5: Find the slope of the normal at the point \( (6, -12) \)
Again, the slope of the normal is the negative reciprocal of the slope of the tangent. If the slope of the tangent \( m_1' = -1 \), then the slope of the normal \( m_2' \) is:

\[ m_2' = -\frac{1}{m_1'} = 1 \]

### Step 6: Find the angle between the two normals
Let \( m_2 = -1 \) (slope of normal at \( (6, 12) \)) and \( m_2' = 1 \) (slope of normal at \( (6, -12) \)). The formula for the angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \) is given by:

\[ \tan \theta = \left| \frac{m_2 - m_2'}{1 + m_2 m_2'} \right| \]

Substituting the values:

\[ \tan \theta = \left| \frac{-1 - 1}{1 + (-1)(1)} \right| = \left| \frac{-2}{1 - 1} \right| \]

Since the denominator becomes zero, \( \tan \theta \) approaches infinity, indicating that \( \theta = \frac{\pi}{2} \) or \( 90^\circ \).

### Final Answer
The angle between the normals at the points \( (6, 12) \) and \( (6, -12) \) is \( 90^\circ \).

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