The normal to the parabola `y^(2)=8x` at the point (2, 4) meets the parabola again at eh point

Options

A. `(-18, -12)`
B. `(-18, 12)`
C. `(18, 12)`
D. `(18, -12)`

Answer

To solve the problem, we need to find the point where the normal to the parabola \( y^2 = 8x \) at the point \( (2, 4) \) intersects the parabola again. Let's go through the steps systematically.

### Step 1: Find the slope of the tangent at the point (2, 4)

The equation of the parabola is given by:
\[ y^2 = 8x \]

Differentiating both sides with respect to \( x \):
\[ 2y \frac{dy}{dx} = 8 \]
\[ \frac{dy}{dx} = \frac{8}{2y} = \frac{4}{y} \]

Now, substituting \( y = 4 \) (the y-coordinate of the point (2, 4)):
\[ \frac{dy}{dx} = \frac{4}{4} = 1 \]

### Step 2: Find the slope of the normal

The slope of the normal \( m_2 \) is the negative reciprocal of the slope of the tangent \( m_1 \):
\[ m_1 = 1 \implies m_2 = -1 \]

### Step 3: Write the equation of the normal line

Using the point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
Substituting \( (x_1, y_1) = (2, 4) \) and \( m = -1 \):
\[ y - 4 = -1(x - 2) \]
\[ y - 4 = -x + 2 \]
\[ y = -x + 6 \]

### Step 4: Substitute the normal line equation into the parabola equation

We need to find where this line intersects the parabola again. Substitute \( y = -x + 6 \) into the parabola equation \( y^2 = 8x \):
\[ (-x + 6)^2 = 8x \]
Expanding the left side:
\[ x^2 - 12x + 36 = 8x \]
Rearranging gives:
\[ x^2 - 20x + 36 = 0 \]

### Step 5: Solve the quadratic equation

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -20, c = 36 \):
\[ x = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 1 \cdot 36}}{2 \cdot 1} \]
\[ x = \frac{20 \pm \sqrt{400 - 144}}{2} \]
\[ x = \frac{20 \pm \sqrt{256}}{2} \]
\[ x = \frac{20 \pm 16}{2} \]
Calculating the two possible values:
1. \( x = \frac{36}{2} = 18 \)
2. \( x = \frac{4}{2} = 2 \)

### Step 6: Find the corresponding y-coordinates

For \( x = 18 \):
\[ y = -18 + 6 = -12 \]

Thus, the point where the normal intersects the parabola again is \( (18, -12) \).

### Final Answer
The normal to the parabola \( y^2 = 8x \) at the point \( (2, 4) \) meets the parabola again at the point \( (18, -12) \).

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