The area bounded by the curve `y^(2)=x` and the ordinate `x=36` is divided in the ratio `1:7` by the ordinate x=a. Then a=

Options

A. 8
B. 9
C. 7
D. 0

Answer

To solve the problem, we need to find the value of \( a \) such that the area bounded by the curve \( y^2 = x \) and the ordinates \( x = 0 \) and \( x = 36 \) is divided in the ratio \( 1:7 \) by the ordinate \( x = a \).

### Step-by-Step Solution:

1. **Identify the Curve and the Area**:
The curve given is \( y^2 = x \). This implies \( y = \sqrt{x} \). We need to find the area under this curve from \( x = 0 \) to \( x = 36 \).

2. **Calculate the Total Area**:
The total area \( A \) under the curve from \( x = 0 \) to \( x = 36 \) can be calculated using the integral:
\[ A = \int_0^{36} \sqrt{x} \, dx \]
The integral of \( \sqrt{x} \) is:
\[ \int \sqrt{x} \, dx = \frac{2}{3} x^{3/2} \]
Evaluating from 0 to 36:
\[ A = \left[ \frac{2}{3} x^{3/2} \right]_0^{36} = \frac{2}{3} (36)^{3/2} - \frac{2}{3} (0)^{3/2} = \frac{2}{3} \cdot 216 = 144 \]

3. **Divide the Area**:
Let the area from \( x = 0 \) to \( x = a \) be \( A_1 \) and the area from \( x = a \) to \( x = 36 \) be \( A_2 \). We know that:
\[ \frac{A_1}{A_2} = \frac{1}{7} \]
This implies:
\[ A_1 = \frac{1}{8} \cdot 144 = 18 \quad \text{and} \quad A_2 = \frac{7}{8} \cdot 144 = 126 \]

4. **Calculate Area \( A_1 \)**:
The area \( A_1 \) from \( x = 0 \) to \( x = a \) is:
\[ A_1 = \int_0^a \sqrt{x} \, dx = \frac{2}{3} a^{3/2} \]
Setting this equal to 18:
\[ \frac{2}{3} a^{3/2} = 18 \]
Multiplying both sides by \( \frac{3}{2} \):
\[ a^{3/2} = 27 \]

5. **Solve for \( a \)**:
To find \( a \), we raise both sides to the power of \( \frac{2}{3} \):
\[ a = 27^{\frac{2}{3}} = (3^3)^{\frac{2}{3}} = 3^2 = 9 \]

Thus, the value of \( a \) is \( 9 \).

### Final Answer:
\[ \boxed{9} \]