Let P be the point (other than the origin) of intersection of the curves `y^(2)=4ax and ay^(2) = 4x^(3)` such that the normals to the two curves meet x-axis at `G_(1) and G_(2)` respectively. Then, `G_(1)G_(2)=`

Options

A. 2a
B. 4a
C. a
D. none of these

Answer

To solve the problem, we need to find the distance \( G_1 G_2 \) between the points where the normals to the curves \( y^2 = 4ax \) and \( ay^2 = 4x^3 \) intersect the x-axis.

### Step-by-Step Solution:

1. **Find the Point of Intersection of the Curves:**
The curves given are:
\[ y^2 = 4ax \quad \text{(1)} \]
\[ ay^2 = 4x^3 \quad \text{(2)} \]
From (1), we can express \( y^2 \) as \( 4ax \). Substitute this into (2):
\[ a(4ax) = 4x^3 \]
Simplifying this gives:
\[ 4a^2x = 4x^3 \]
Dividing both sides by 4:
\[ a^2x = x^3 \]
Rearranging gives:
\[ x^3 - a^2x = 0 \]
Factoring out \( x \):
\[ x(x^2 - a^2) = 0 \]
This gives us \( x = 0 \) or \( x = \pm a \). Since we are looking for the point other than the origin, we take \( x = a \).

2. **Find the Corresponding y-values:**
Substitute \( x = a \) back into equation (1):
\[ y^2 = 4a(a) = 4a^2 \]
Thus, \( y = \pm 2a \). Therefore, the points of intersection are \( P(a, 2a) \) and \( P(a, -2a) \).

3. **Find the Slopes of the Normals:**
- For the curve \( y^2 = 4ax \):
\[ \frac{dy}{dx} = \frac{2a}{y} \]
At \( P(a, 2a) \):
\[ \frac{dy}{dx} = \frac{2a}{2a} = 1 \]
The slope of the normal is:
\[ m_1 = -1 \]

- For the curve \( ay^2 = 4x^3 \):
\[ \frac{dy}{dx} = \frac{12x^2}{2ay} = \frac{6x^2}{ay} \]
At \( P(a, 2a) \):
\[ \frac{dy}{dx} = \frac{6a^2}{2a^2} = 3 \]
The slope of the normal is:
\[ m_2 = -\frac{1}{3} \]

4. **Find the Equations of the Normals:**
- For the first curve at \( P(a, 2a) \):
\[ y - 2a = -1(x - a) \implies y = -x + 3a \]
- For the second curve at \( P(a, 2a) \):
\[ y - 2a = -\frac{1}{3}(x - a) \implies 3(y - 2a) = -1(x - a) \]
Simplifying gives:
\[ 3y - 6a = -x + a \implies x + 3y = 7a \]

5. **Find the Points \( G_1 \) and \( G_2 \) where Normals Meet the x-axis:**
- For \( G_1 \) (where \( y = 0 \)):
\[ 0 = -x + 3a \implies x = 3a \implies G_1(3a, 0) \]
- For \( G_2 \) (where \( y = 0 \)):
\[ 0 + 3y = 7a \implies x = 7a \implies G_2(7a, 0) \]

6. **Calculate the Distance \( G_1 G_2 \):**
The distance \( G_1 G_2 \) is:
\[ G_1 G_2 = |7a - 3a| = 4a \]

### Final Answer:
\[ G_1 G_2 = 4a \]