If `PG_1` and `PG_2` be the normals to the curves `y^2 = 4ax` and `ay^2 = 4x^3` at a common point other than origin meeting x-axis in `G_1` and `G_2` , then `G_1G_2` =

Options

A. 2a
B. 4a
C. 6a
D. none

Answer

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

### Step 1: Find the common point of the curves
The first curve is given by:
\[ y^2 = 4ax \]
The second curve is given by:
\[ ay^2 = 4x^3 \]
Substituting \( y^2 = 4ax \) into the second equation:
\[ a(4ax) = 4x^3 \implies 4a^2x = 4x^3 \implies a^2x = x^3 \]
This gives us:
\[ x^3 - a^2x = 0 \implies x(x^2 - a^2) = 0 \]
Thus, \( x = 0 \) or \( x = a \) or \( x = -a \). Since we are looking for a common point other than the origin, we take \( x = a \).

Substituting \( x = a \) back into the first curve to find \( y \):
\[ y^2 = 4a(a) \implies y^2 = 4a^2 \implies y = 2a \text{ (taking the positive root)} \]
So, the common point is \( (a, 2a) \).

### Step 2: Find the equation of the normal to the first curve
For the curve \( y^2 = 4ax \):
1. Differentiate to find \( \frac{dy}{dx} \):
\[ 2y \frac{dy}{dx} = 4a \implies \frac{dy}{dx} = \frac{4a}{2y} = \frac{4a}{4a} = 1 \]
2. The slope of the normal \( m_1 \) is:
\[ m_1 = -\frac{1}{1} = -1 \]
3. The equation of the normal at point \( (a, 2a) \):
\[ y - 2a = -1(x - a) \implies y - 2a = -x + a \implies y = -x + 3a \]

### Step 3: Find the x-intercept \( G_1 \)
To find \( G_1 \), set \( y = 0 \):
\[ 0 = -x + 3a \implies x = 3a \]
Thus, \( G_1 = (3a, 0) \).

### Step 4: Find the equation of the normal to the second curve
For the curve \( ay^2 = 4x^3 \):
1. Differentiate to find \( \frac{dy}{dx} \):
\[ 2ay \frac{dy}{dx} = 12x^2 \implies \frac{dy}{dx} = \frac{12x^2}{2ay} = \frac{6x^2}{ay} \]
2. At the point \( (a, 2a) \):
\[ \frac{dy}{dx} = \frac{6a^2}{2a^2} = 3 \]
3. The slope of the normal \( m_2 \) is:
\[ m_2 = -\frac{1}{3} \]
4. The equation of the normal at point \( (a, 2a) \):
\[ y - 2a = -\frac{1}{3}(x - a) \implies 3(y - 2a) = -x + a \implies 3y - 6a = -x + a \implies x + 3y = 7a \]

### Step 5: Find the x-intercept \( G_2 \)
To find \( G_2 \), set \( y = 0 \):
\[ x + 3(0) = 7a \implies x = 7a \]
Thus, \( G_2 = (7a, 0) \).

### Step 6: Find the distance \( G_1G_2 \)
The distance \( G_1G_2 \) is given by:
\[ G_1G_2 = |7a - 3a| = |4a| = 4a \]

### Final Answer
Thus, the distance \( G_1G_2 \) is \( 4a \).

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