If the area enclosed by `y^(2) = 4ax` and the line y=ax is `(1)/(3)` sq. unit, then the roots of the equation `x^(2) + 2x = a`, are

Options

A. -4 and 2
B. 2 and 4
C. -2 and -4
D. 8 and -8

Answer

To solve the problem, we need to find the roots of the equation \( x^2 + 2x = a \) given that the area enclosed by the curve \( y^2 = 4ax \) and the line \( y = ax \) is \( \frac{1}{3} \) square units.

### Step 1: Find the points of intersection

The equations we have are:
1. \( y^2 = 4ax \) (parabola)
2. \( y = ax \) (line)

Substituting \( y = ax \) into \( y^2 = 4ax \):

\[ (ax)^2 = 4ax \]
\[ a^2x^2 = 4ax \]
\[ a^2x^2 - 4ax = 0 \]
Factoring out \( ax \):

\[ ax(a - 4) = 0 \]

This gives us two solutions:
1. \( x = 0 \)
2. \( a - 4 = 0 \) or \( a = 4 \)

### Step 2: Find the corresponding \( y \) values

For \( x = 0 \):
\[ y = a(0) = 0 \quad \Rightarrow \quad (0, 0) \]

For \( a = 4 \):
\[ y = 4x \quad \Rightarrow \quad y = 4(1) = 4 \quad \Rightarrow \quad (1, 4) \]

So the points of intersection are \( (0, 0) \) and \( (4, 4) \).

### Step 3: Set up the integral for the area

The area \( A \) between the curves from \( x = 0 \) to \( x = 4 \) is given by:

\[ A = \int_0^4 (y_{\text{top}} - y_{\text{bottom}}) \, dx \]

Here, \( y_{\text{top}} = ax \) and \( y_{\text{bottom}} = \frac{y^2}{4a} \) (from the parabola).

Thus, the area becomes:

\[ A = \int_0^4 \left(ax - \frac{y^2}{4a}\right) \, dx \]

### Step 4: Calculate the area

Substituting \( y = ax \) into the area equation:

\[ A = \int_0^4 \left(ax - \frac{(ax)^2}{4a}\right) \, dx \]

This simplifies to:

\[ A = \int_0^4 \left(ax - \frac{a^2x^2}{4a}\right) \, dx = \int_0^4 \left(ax - \frac{ax^2}{4}\right) \, dx \]

Now, we can integrate:

\[ A = \int_0^4 \left(ax - \frac{ax^2}{4}\right) \, dx = \left[\frac{ax^2}{2} - \frac{ax^3}{12}\right]_0^4 \]

Calculating this:

\[ = \left[\frac{a(4^2)}{2} - \frac{a(4^3)}{12}\right] - \left[0\right] \]
\[ = \left[\frac{16a}{2} - \frac{64a}{12}\right] \]
\[ = 8a - \frac{16a}{3} \]
\[ = \frac{24a}{3} - \frac{16a}{3} = \frac{8a}{3} \]

### Step 5: Set the area equal to \( \frac{1}{3} \)

Given that the area is \( \frac{1}{3} \):

\[ \frac{8a}{3} = \frac{1}{3} \]

Multiplying both sides by 3:

\[ 8a = 1 \quad \Rightarrow \quad a = \frac{1}{8} \]

### Step 6: Find the roots of the equation \( x^2 + 2x = a \)

Substituting \( a \):

\[ x^2 + 2x = \frac{1}{8} \]
\[ x^2 + 2x - \frac{1}{8} = 0 \]

Multiplying through by 8 to eliminate the fraction:

\[ 8x^2 + 16x - 1 = 0 \]

### Step 7: Use the quadratic formula to find the roots

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 8, b = 16, c = -1 \):

\[ x = \frac{-16 \pm \sqrt{16^2 - 4 \cdot 8 \cdot (-1)}}{2 \cdot 8} \]
\[ = \frac{-16 \pm \sqrt{256 + 32}}{16} \]
\[ = \frac{-16 \pm \sqrt{288}}{16} \]
\[ = \frac{-16 \pm 12\sqrt{2}}{16} \]
\[ = \frac{-4 \pm 3\sqrt{2}}{4} \]

Thus, the roots are:

\[ x = -1 + \frac{3\sqrt{2}}{4}, \quad x = -1 - \frac{3\sqrt{2}}{4} \]

### Final Answer
The roots of the equation \( x^2 + 2x = a \) are:

\[ x = -1 + \frac{3\sqrt{2}}{4}, \quad x = -1 - \frac{3\sqrt{2}}{4} \]