Let A, be the area of the parabola `y2 = 4ax` lying between vertex and latus rectum and `A_(2)`, be the area between latus rectum and double ordinate `x=2a`. Then `A_(1)/A_(2)=`.

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: Define the areas \( A_1 \) and \( A_2 \)

1. **Area \( A_1 \)**: This is the area of the parabola \( y^2 = 4ax \) lying between the vertex (origin) and the latus rectum (line \( x = a \)).
2. **Area \( A_2 \)**: This is the area between the latus rectum (line \( x = a \)) and the double ordinate \( x = 2a \).

### Step 2: Calculate \( A_1 \)

To find \( A_1 \), we will use integration. The area under the curve from \( x = 0 \) to \( x = a \) can be calculated as follows:

\[ A_1 = 2 \int_0^a y \, dx \]

From the equation of the parabola \( y^2 = 4ax \), we have:

\[ y = \sqrt{4ax} = 2\sqrt{ax} \]

Thus, substituting \( y \) into the integral, we have:

\[ A_1 = 2 \int_0^a 2\sqrt{ax} \, dx = 4 \int_0^a \sqrt{ax} \, dx \]

Now, we can take \( \sqrt{a} \) out of the integral:

\[ A_1 = 4\sqrt{a} \int_0^a \sqrt{x} \, dx \]

The integral of \( \sqrt{x} \) is:

\[ \int \sqrt{x} \, dx = \frac{2}{3} x^{3/2} \]

Now, we evaluate the definite integral from \( 0 \) to \( a \):

\[ \int_0^a \sqrt{x} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_0^a = \frac{2}{3} a^{3/2} \]

Substituting this back into the expression for \( A_1 \):

\[ A_1 = 4\sqrt{a} \cdot \frac{2}{3} a^{3/2} = \frac{8}{3} a^2 \]

### Step 3: Calculate \( A_2 \)

Next, we calculate \( A_2 \) using a similar approach:

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

Again, using \( y = 2\sqrt{ax} \):

\[ A_2 = 2 \int_a^{2a} 2\sqrt{ax} \, dx = 4 \int_a^{2a} \sqrt{ax} \, dx \]

Taking \( \sqrt{a} \) out:

\[ A_2 = 4\sqrt{a} \int_a^{2a} \sqrt{x} \, dx \]

Now we evaluate the integral:

\[ \int \sqrt{x} \, dx = \frac{2}{3} x^{3/2} \]

Evaluating from \( a \) to \( 2a \):

\[ \int_a^{2a} \sqrt{x} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_a^{2a} = \frac{2}{3} \left( (2a)^{3/2} - a^{3/2} \right) \]

Calculating \( (2a)^{3/2} \):

\[ (2a)^{3/2} = 2^{3/2} a^{3/2} = 2\sqrt{2} a^{3/2} \]

Thus,

\[ \int_a^{2a} \sqrt{x} \, dx = \frac{2}{3} \left( 2\sqrt{2} a^{3/2} - a^{3/2} \right) = \frac{2}{3} (2\sqrt{2} - 1) a^{3/2} \]

Substituting back into \( A_2 \):

\[ A_2 = 4\sqrt{a} \cdot \frac{2}{3} (2\sqrt{2} - 1) a^{3/2} = \frac{8}{3} (2\sqrt{2} - 1) a^2 \]

### Step 4: Calculate the ratio \( \frac{A_1}{A_2} \)

Now we can find the ratio:

\[ \frac{A_1}{A_2} = \frac{\frac{8}{3} a^2}{\frac{8}{3} (2\sqrt{2} - 1) a^2} = \frac{1}{2\sqrt{2} - 1} \]

### Final Answer

Thus, the final result is:

\[ \frac{A_1}{A_2} = \frac{1}{2\sqrt{2} - 1} \]