The area of the region enclosed between parabola `y^(2)=x` and the line `y=mx` is `(1)/(48)`. Then the value of m is

Options

A. `-2`
B. `-1`
C. 1
D. 2

Answer

To find the value of \( m \) such that the area of the region enclosed between the parabola \( y^2 = x \) and the line \( y = mx \) is \( \frac{1}{48} \), we can follow these steps:

### Step 1: Find the points of intersection
The parabola is given by \( y^2 = x \) and the line by \( y = mx \). To find the points of intersection, we substitute \( y = mx \) into the equation of the parabola:

\[ (mx)^2 = x \]

This simplifies to:

\[ m^2 x^2 - x = 0 \]

Factoring out \( x \):

\[ x(m^2 x - 1) = 0 \]

This gives us two solutions:

1. \( x = 0 \) (the origin)
2. \( m^2 x - 1 = 0 \) which gives \( x = \frac{1}{m^2} \)

Now, substituting \( x = \frac{1}{m^2} \) back into \( y = mx \):

\[ y = m \left(\frac{1}{m^2}\right) = \frac{1}{m} \]

Thus, the points of intersection are \( (0, 0) \) and \( \left(\frac{1}{m^2}, \frac{1}{m}\right) \).

### Step 2: Calculate the area of the triangle formed
The area of triangle \( ABC \) formed by the points \( A\left(\frac{1}{m^2}, \frac{1}{m}\right) \), \( B(0, 0) \), and \( C\left(\frac{1}{m^2}, 0\right) \) can be calculated using the formula for the area of a triangle:

\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]

Here, the base is \( \frac{1}{m^2} \) and the height is \( \frac{1}{m} \):

\[ \text{Area}_{\text{triangle}} = \frac{1}{2} \times \frac{1}{m^2} \times \frac{1}{m} = \frac{1}{2m^3} \]

### Step 3: Calculate the area under the parabola
Next, we calculate the area under the parabola from \( x = 0 \) to \( x = \frac{1}{m^2} \):

\[ \text{Area}_{\text{parabola}} = \int_0^{\frac{1}{m^2}} \sqrt{x} \, dx \]

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

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

Evaluating from \( 0 \) to \( \frac{1}{m^2} \):

\[ \text{Area}_{\text{parabola}} = \left[\frac{2}{3} x^{\frac{3}{2}}\right]_0^{\frac{1}{m^2}} = \frac{2}{3} \left(\frac{1}{m^2}\right)^{\frac{3}{2}} = \frac{2}{3} \cdot \frac{1}{m^3} \]

### Step 4: Calculate the area between the curves
The area between the curves is the area of the triangle minus the area under the parabola:

\[ \text{Area}_{\text{between}} = \text{Area}_{\text{triangle}} - \text{Area}_{\text{parabola}} = \frac{1}{2m^3} - \frac{2}{3m^3} \]

Finding a common denominator (which is \( 6m^3 \)):

\[ \text{Area}_{\text{between}} = \frac{3}{6m^3} - \frac{4}{6m^3} = \frac{-1}{6m^3} \]

### Step 5: Set the area equal to \( \frac{1}{48} \)
According to the problem, this area is equal to \( \frac{1}{48} \):

\[ \frac{-1}{6m^3} = \frac{1}{48} \]

Multiplying both sides by \( -1 \):

\[ \frac{1}{6m^3} = -\frac{1}{48} \]

Cross-multiplying gives:

\[ 48 = 6m^3 \]

Dividing both sides by 6:

\[ m^3 = 8 \]

Taking the cube root:

\[ m = 2 \quad \text{or} \quad m = -2 \]

### Final Answer
The possible values of \( m \) are \( 2 \) and \( -2 \).