If `A_1` is the area of the parabola `y^2=4 ax` lying between vertex and the latusrectum and `A_2` is the area between the latusrectum and the double ordinate `x=2 a`, then `A_1/A_2` is equal to

Options

A. `2sqrt(2)-1`
B. `(2sqrt(2)+1)//7`
C. `(2sqrt(2)-1)//7`
D. none of these

Answer

To solve the problem, we need to find the areas \( A_1 \) and \( A_2 \) as defined in the question, and then compute the ratio \( \frac{A_1}{A_2} \).

### Step 1: Understand the Parabola
The equation of the parabola is given as \( y^2 = 4ax \). The vertex of the parabola is at the origin (0,0), and the latus rectum is a vertical line that passes through the focus of the parabola, which is at the point \( (a, 0) \).

### Step 2: Determine the Area \( A_1 \)
The area \( A_1 \) is the area between the vertex (0,0) and the latus rectum (which is at \( x = a \)). To find this area, we will integrate the function representing the parabola from \( x = 0 \) to \( x = a \).

The equation can be rewritten for \( y \):
\[ y = 2\sqrt{ax} \]

Now, we calculate the area \( A_1 \):
\[ A_1 = \int_0^a 2\sqrt{ax} \, dx \]

### Step 3: Solve the Integral for \( A_1 \)
To solve the integral:
\[ A_1 = 2\sqrt{a} \int_0^a \sqrt{x} \, dx \]

The integral \( \int \sqrt{x} \, dx \) is:
\[ \int \sqrt{x} \, dx = \frac{2}{3} x^{3/2} \]

Thus,
\[ A_1 = 2\sqrt{a} \left[ \frac{2}{3} x^{3/2} \right]_0^a = 2\sqrt{a} \cdot \frac{2}{3} a^{3/2} = \frac{4}{3} a^2 \]

### Step 4: Determine the Area \( A_2 \)
The area \( A_2 \) is the area between the latus rectum (at \( x = a \)) and the double ordinate \( x = 2a \). We will integrate the same function from \( x = a \) to \( x = 2a \).

\[ A_2 = \int_a^{2a} 2\sqrt{ax} \, dx \]

### Step 5: Solve the Integral for \( A_2 \)
Again, we can factor out \( 2\sqrt{a} \):
\[ A_2 = 2\sqrt{a} \int_a^{2a} \sqrt{x} \, dx \]

Calculating the integral:
\[ A_2 = 2\sqrt{a} \left[ \frac{2}{3} x^{3/2} \right]_a^{2a} = 2\sqrt{a} \left( \frac{2}{3} (2a)^{3/2} - \frac{2}{3} a^{3/2} \right) \]

Calculating \( (2a)^{3/2} = 2^{3/2} a^{3/2} = 2\sqrt{2} a^{3/2} \):
\[ A_2 = 2\sqrt{a} \cdot \frac{2}{3} \left( 2\sqrt{2} a^{3/2} - a^{3/2} \right) = \frac{4\sqrt{a}}{3} \left( (2\sqrt{2} - 1) a^{3/2} \right) \]
\[ A_2 = \frac{4}{3} a^2 (2\sqrt{2} - 1) \]

### Step 6: Calculate the Ratio \( \frac{A_1}{A_2} \)
Now we have:
\[ A_1 = \frac{4}{3} a^2 \]
\[ A_2 = \frac{4}{3} a^2 (2\sqrt{2} - 1) \]

Thus, the ratio is:
\[ \frac{A_1}{A_2} = \frac{\frac{4}{3} a^2}{\frac{4}{3} a^2 (2\sqrt{2} - 1)} = \frac{1}{2\sqrt{2} - 1} \]

### Step 7: Rationalize the Denominator
To rationalize \( \frac{1}{2\sqrt{2} - 1} \):
\[ \frac{1}{2\sqrt{2} - 1} \cdot \frac{2\sqrt{2} + 1}{2\sqrt{2} + 1} = \frac{2\sqrt{2} + 1}{(2\sqrt{2})^2 - 1^2} = \frac{2\sqrt{2} + 1}{8 - 1} = \frac{2\sqrt{2} + 1}{7} \]

### Final Answer
Thus, the final result is:
\[ \frac{A_1}{A_2} = \frac{2\sqrt{2} + 1}{7} \]