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

Options

A. (-18, -12)
B. (-28, 12)
C. (18, 12)
D. (18, -12)

Answer

To find the point where the normal to the parabola \( y^2 = 8ax \) at the point \( (2, 4) \) meets the parabola again, we can follow these steps:

### Step 1: Verify that the point lies on the parabola
The given parabola is \( y^2 = 8ax \). We need to substitute the point \( (2, 4) \) into the equation to find the value of \( a \).

\[ 4^2 = 8a \cdot 2 \]
\[ 16 = 16a \]
\[ a = 1 \]

### Step 2: Write the equation of the parabola
Now that we have \( a = 1 \), we can write the equation of the parabola:

\[ y^2 = 8x \]

### Step 3: Find the parameter \( t_1 \)
The point \( (2, 4) \) can be expressed in terms of the parameter \( t_1 \) for the parabola \( y^2 = 8x \). The parametric coordinates are given by:

\[ (x, y) = (2a t_1^2, 2a t_1) \]

Substituting \( a = 1 \):

\[ (2, 4) = (2t_1^2, 2t_1) \]

From \( 2t_1^2 = 2 \):

\[ t_1^2 = 1 \implies t_1 = 1 \text{ or } t_1 = -1 \]

From \( 2t_1 = 4 \):

\[ t_1 = 2 \]

Since both equations must hold, we take \( t_1 = 2 \).

### Step 4: Use the normal condition to find \( t_2 \)
The relation between \( t_1 \) and \( t_2 \) for the normal to the parabola is given by:

\[ t_2 = -t_1 - \frac{2}{t_1} \]

Substituting \( t_1 = 2 \):

\[ t_2 = -2 - \frac{2}{2} = -2 - 1 = -3 \]

### Step 5: Find the coordinates of the point where the normal intersects the parabola again
Now we can find the coordinates of the point corresponding to \( t_2 \):

\[ (x, y) = (2a t_2^2, 2a t_2) \]

Substituting \( a = 1 \) and \( t_2 = -3 \):

\[ x = 2 \cdot 1 \cdot (-3)^2 = 2 \cdot 1 \cdot 9 = 18 \]
\[ y = 2 \cdot 1 \cdot (-3) = -6 \]

Thus, the coordinates of the point where the normal meets the parabola again are \( (18, -6) \).

### Final Answer
The point where the normal to the parabola \( y^2 = 8x \) at the point \( (2, 4) \) meets the parabola again is \( (18, -6) \).

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