If the normals to the parabola `y^2=4a x` at the ends of the latus rectum meet the parabola at `Qa n dQ^(prime),` then `QQ '` is `10 a` (b) `4a` (c) `20 c` (d) `12 a`

Options

A. 10a
B. 4a
C. 20a
D. 12a

Explanation

To solve the problem, we need to find the length \( QQ' \) where \( Q \) and \( Q' \) are the points where the normals at the ends of the latus rectum of the parabola \( y^2 = 4ax \) intersect the parabola again.

### Step-by-Step Solution:

1. **Identify the ends of the latus rectum**:
The latus rectum of the parabola \( y^2 = 4ax \) has endpoints at \( (a, 2a) \) and \( (a, -2a) \).

2. **Parameterize the points**:
Let \( t_1 \) be the parameter for the point \( (a, 2a) \) and \( t_2 \) for the point \( (a, -2a) \). For the parabola, the coordinates in terms of the parameter \( t \) are given by:
\[ (at^2, 2at) \]
Therefore, for \( t_1 = 1 \) (for the point \( (a, 2a) \)):
\[ (a(1^2), 2a(1)) = (a, 2a) \]
For \( t_2 = -1 \) (for the point \( (a, -2a) \)):
\[ (a(-1^2), 2a(-1)) = (a, -2a) \]

3. **Find the normals at these points**:
The slope of the tangent at a point \( (at^2, 2at) \) is given by \( \frac{dy}{dx} = \frac{2a}{2at} = \frac{1}{t} \). Therefore, the slope of the normal is \( -t \).

- For \( t_1 = 1 \):
The equation of the normal at \( (a, 2a) \) is:
\[ y - 2a = -1(x - a) \implies y = -x + 3a \]

- For \( t_2 = -1 \):
The equation of the normal at \( (a, -2a) \) is:
\[ y + 2a = 1(x - a) \implies y = x - 3a \]

4. **Find the intersection points \( Q \) and \( Q' \)**:
To find the intersection of the normal \( y = -x + 3a \) with the parabola \( y^2 = 4ax \):
Substitute \( y = -x + 3a \) into \( y^2 = 4ax \):
\[ (-x + 3a)^2 = 4ax \]
Expanding gives:
\[ x^2 - 6ax + 9a^2 = 4ax \implies x^2 - 10ax + 9a^2 = 0 \]
Using the quadratic formula:
\[ x = \frac{10a \pm \sqrt{(10a)^2 - 4 \cdot 1 \cdot 9a^2}}{2 \cdot 1} = \frac{10a \pm \sqrt{100a^2 - 36a^2}}{2} = \frac{10a \pm \sqrt{64a^2}}{2} = \frac{10a \pm 8a}{2} \]
This gives:
\[ x_1 = 9a, \quad x_2 = a \]

For \( x_1 = 9a \):
\[ y_1 = -9a + 3a = -6a \quad \text{(Point Q)} \]
For \( x_2 = a \):
\[ y_2 = -a + 3a = 2a \quad \text{(Point Q')} \]

5. **Calculate the distance \( QQ' \)**:
The points \( Q \) and \( Q' \) are \( (9a, -6a) \) and \( (9a, 6a) \) respectively. The distance \( QQ' \) is given by:
\[ QQ' = |y_2 - y_1| = |6a - (-6a)| = |6a + 6a| = |12a| = 12a \]

Thus, the length \( QQ' \) is \( 12a \).

### Final Answer:
The correct option is (d) \( 12a \).