If a line `x+ y =1` cut the parabola `y^2 = 4ax` in points A and B and normals drawn at A and B meet at C. The normals to the parabola from C other than above two meets the parabola in D, then point D is : (A) `(a,a)` (B) `(2a,2a)` (C) `(3a,3a)` (D) `(4a,4a)`

Options

A. (a,2a)
B. `((4am)/l^2,(4a)/l)`
C. `((2am^2)/l^2,(2a)/l)`
D. `((4am^2)/l^2,(4am)/l)`

Answer

To solve the problem step by step, we will follow the transcript's outline and derive the coordinates of point D.

### Step 1: Find the intersection points A and B of the line and the parabola.

We have the line given by the equation:
\[ x + y = 1 \]
This can be rewritten as:
\[ y = 1 - x \]

The parabola is given by:
\[ y^2 = 4ax \]

Substituting the expression for \( y \) from the line into the parabola's equation:
\[ (1 - x)^2 = 4ax \]
Expanding this:
\[ 1 - 2x + x^2 = 4ax \]
Rearranging gives:
\[ x^2 - (4a + 2)x + 1 = 0 \]

### Step 2: Solve the quadratic equation for x.

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1 \), \( b = -(4a + 2) \), and \( c = 1 \).

Calculating the discriminant:
\[ b^2 - 4ac = (4a + 2)^2 - 4 \cdot 1 \cdot 1 = 16a^2 + 16a + 4 - 4 = 16a^2 + 16a \]

Now substituting into the quadratic formula:
\[ x = \frac{4a + 2 \pm \sqrt{16a^2 + 16a}}{2} \]
Simplifying:
\[ x = \frac{4a + 2 \pm 4\sqrt{a(a + 1)}}{2} = 2a + 1 \pm 2\sqrt{a(a + 1)} \]
Thus, we have two x-coordinates for points A and B:
\[ x_1 = 2a + 1 + 2\sqrt{a(a + 1)}, \quad x_2 = 2a + 1 - 2\sqrt{a(a + 1)} \]

### Step 3: Find the corresponding y-coordinates.

Using \( y = 1 - x \):
For \( x_1 \):
\[ y_1 = 1 - (2a + 1 + 2\sqrt{a(a + 1)}) = -2a - 2\sqrt{a(a + 1)} \]
For \( x_2 \):
\[ y_2 = 1 - (2a + 1 - 2\sqrt{a(a + 1)}) = -2a + 2\sqrt{a(a + 1)} \]

### Step 4: Find the coordinates of point C.

The normals at points A and B meet at point C. The property of the parabola states that the sum of the y-coordinates of points A, B, and D is zero:
\[ y_1 + y_2 + y_D = 0 \]
Substituting the y-coordinates:
\[ (-2a - 2\sqrt{a(a + 1)}) + (-2a + 2\sqrt{a(a + 1)}) + y_D = 0 \]
This simplifies to:
\[ -4a + y_D = 0 \implies y_D = 4a \]

### Step 5: Find the x-coordinate of point D.

Since point D lies on the parabola \( y^2 = 4ax \):
\[ (4a)^2 = 4a \cdot x_D \implies 16a^2 = 4ax_D \implies x_D = 4a \]

### Conclusion: Coordinates of point D.

Thus, the coordinates of point D are:
\[ D = (4a, 4a) \]

### Final Answer:
The correct option is (D) \( (4a, 4a) \).