Find the area of the figure enclosed by the curve `5x^2+6x y+2y^2+7x+6y+6=0.` (in Sq. unit)

Options

A. `pi/4`
B. `pi/2`
C. `pi`
D. `2pi`

Answer

To find the area enclosed by the curve given by the equation \(5x^2 + 6xy + 2y^2 + 7x + 6y + 6 = 0\), we can follow these steps:

### Step 1: Rewrite the Equation
We start with the equation:
\[ 5x^2 + 6xy + 2y^2 + 7x + 6y + 6 = 0 \]
This is a quadratic equation in terms of \(y\). We can rearrange it to express \(y\) in terms of \(x\).

### Step 2: Identify Coefficients
The general form of a quadratic equation in \(y\) is:
\[ Ay^2 + By + C = 0 \]
In our case, we can identify:
- \(A = 2\)
- \(B = 6x + 6\)
- \(C = 5x^2 + 7x + 6\)

### Step 3: Calculate the Discriminant
To find the area enclosed by the curve, we need to ensure that the quadratic in \(y\) has real solutions. The discriminant \(D\) of a quadratic equation \(Ay^2 + By + C = 0\) is given by:
\[ D = B^2 - 4AC \]
Substituting our values:
\[ D = (6x + 6)^2 - 4 \cdot 2 \cdot (5x^2 + 7x + 6) \]
Calculating \(D\):
\[ D = (36x^2 + 72x + 36) - (40x^2 + 56x + 48) \]
\[ D = -4x^2 + 16x - 12 \]

### Step 4: Find the Roots of the Discriminant
Setting the discriminant \(D\) to zero to find the points where the curve intersects:
\[ -4x^2 + 16x - 12 = 0 \]
Dividing through by -4:
\[ x^2 - 4x + 3 = 0 \]
Factoring:
\[ (x - 1)(x - 3) = 0 \]
Thus, \(x = 1\) and \(x = 3\).

### Step 5: Find the Area
To find the area enclosed by the curve, we need to integrate the function representing \(y\) over the interval from \(x = 1\) to \(x = 3\).

Using the quadratic formula to find \(y\):
\[ y = \frac{-(6x + 6) \pm \sqrt{D}}{2A} \]
Calculating the two values of \(y\) gives us the upper and lower bounds of the area.

### Step 6: Set Up the Integral
The area \(A\) can be calculated as:
\[ A = \int_{1}^{3} (y_{upper} - y_{lower}) \, dx \]
Where \(y_{upper}\) and \(y_{lower}\) are the two solutions for \(y\) derived from the quadratic formula.

### Step 7: Evaluate the Integral
After substituting the expressions for \(y_{upper}\) and \(y_{lower}\), we can evaluate the integral to find the area.

### Final Answer
After evaluating the integral, we find that the area enclosed by the curve is \(A\) square units.

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