If area bounded by the curves `y^(2)=4axandy=mx` is `a^(2)//3` then the value of m is

Options

A. 1
B. 2
C. 3
D. `(1)/(2)`

Answer

To solve the problem of finding the value of \( m \) such that the area bounded by the curves \( y^2 = 4ax \) and \( y = mx \) is \( \frac{a^2}{3} \), we will follow these steps:

### Step 1: Find the points of intersection
We need to find the points where the curves intersect. We have:
1. \( y^2 = 4ax \) (Equation of the parabola)
2. \( y = mx \) (Equation of the line)

Substituting \( y = mx \) into the parabola's equation:
\[ (mx)^2 = 4ax \]
This simplifies to:
\[ m^2x^2 = 4ax \]
Rearranging gives:
\[ m^2x^2 - 4ax = 0 \]
Factoring out \( x \):
\[ x(m^2x - 4a) = 0 \]
Thus, the solutions are:
\[ x = 0 \quad \text{or} \quad m^2x - 4a = 0 \implies x = \frac{4a}{m^2} \]

### Step 2: Find the corresponding \( y \) values
Using \( x = \frac{4a}{m^2} \) in the line equation \( y = mx \):
\[ y = m \left(\frac{4a}{m^2}\right) = \frac{4a}{m} \]
So the points of intersection are:
\[ (0, 0) \quad \text{and} \quad \left(\frac{4a}{m^2}, \frac{4a}{m}\right) \]

### Step 3: Set up the area integral
The area \( A \) bounded by the curves from \( x = 0 \) to \( x = \frac{4a}{m^2} \) is given by:
\[ A = \int_0^{\frac{4a}{m^2}} \left( \sqrt{4ax} - mx \right) dx \]
Here, \( \sqrt{4ax} \) is the upper curve (parabola) and \( mx \) is the lower curve (line).

### Step 4: Evaluate the integral
Calculating the integral:
\[ A = \int_0^{\frac{4a}{m^2}} \left( \sqrt{4ax} - mx \right) dx \]
The integral can be split into two parts:
\[ A = \int_0^{\frac{4a}{m^2}} \sqrt{4ax} \, dx - \int_0^{\frac{4a}{m^2}} mx \, dx \]

Calculating the first integral:
\[ \int \sqrt{4ax} \, dx = \frac{2}{3} (4a)^{1/2} x^{3/2} = \frac{4\sqrt{a}}{3} x^{3/2} \]
Evaluating from 0 to \( \frac{4a}{m^2} \):
\[ \frac{4\sqrt{a}}{3} \left(\frac{4a}{m^2}\right)^{3/2} = \frac{4\sqrt{a}}{3} \cdot \frac{8a^{3/2}}{m^3} = \frac{32a^2}{3m^3} \]

Calculating the second integral:
\[ \int mx \, dx = \frac{m}{2} x^2 \]
Evaluating from 0 to \( \frac{4a}{m^2} \):
\[ \frac{m}{2} \left(\frac{4a}{m^2}\right)^2 = \frac{m}{2} \cdot \frac{16a^2}{m^4} = \frac{8a^2}{m^3} \]

Putting it all together:
\[ A = \frac{32a^2}{3m^3} - \frac{8a^2}{m^3} = \frac{32a^2 - 24a^2}{3m^3} = \frac{8a^2}{3m^3} \]

### Step 5: Set the area equal to \( \frac{a^2}{3} \)
According to the problem, this area is equal to \( \frac{a^2}{3} \):
\[ \frac{8a^2}{3m^3} = \frac{a^2}{3} \]
Cancelling \( \frac{a^2}{3} \) from both sides (assuming \( a \neq 0 \)):
\[ 8 = m^3 \]
Thus, we find:
\[ m^3 = 8 \implies m = 2 \]

### Final Answer
The value of \( m \) is \( 2 \).

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