The family of curves represented by `(dy)/(dx)=(x^(2)+x+1)/(y^(2)+y+1)` and the family represented by `(dy)/(dx)+(y^(2)+y+1)/(x^(2)+x+1)=0`

Options

A. touch each other
B. are orthogonal
C. are one and the differential
D. none of these

Answer

To determine the relationship between the two families of curves given by the equations:

1. \(\frac{dy}{dx} = \frac{x^2 + x + 1}{y^2 + y + 1}\)
2. \(\frac{dy}{dx} + \frac{y^2 + y + 1}{x^2 + x + 1} = 0\)

we will analyze the equations step by step.

### Step 1: Rewrite the second equation
We start with the second equation and isolate \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = -\frac{y^2 + y + 1}{x^2 + x + 1} \]

### Step 2: Compare the two equations
Now we have two expressions for \(\frac{dy}{dx}\):

1. From the first equation:
\[ \frac{dy}{dx} = \frac{x^2 + x + 1}{y^2 + y + 1} \]

2. From the second equation:
\[ \frac{dy}{dx} = -\frac{y^2 + y + 1}{x^2 + x + 1} \]

### Step 3: Set the two expressions equal
Since both expressions represent \(\frac{dy}{dx}\), we can set them equal to each other:

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

### Step 4: Cross-multiply
Cross-multiplying gives us:

\[ (x^2 + x + 1)^2 = -(y^2 + y + 1)^2 \]

### Step 5: Analyze the equality
The left-hand side \((x^2 + x + 1)^2\) is always non-negative since it is a square. The right-hand side \(-(y^2 + y + 1)^2\) is always non-positive since it is the negative of a square. The only way for these two sides to be equal is if both sides are equal to zero:

\[ (x^2 + x + 1)^2 = 0 \quad \text{and} \quad (y^2 + y + 1)^2 = 0 \]

### Step 6: Solve for conditions
This implies:

1. \(x^2 + x + 1 = 0\) has no real solutions (discriminant \(< 0\)).
2. \(y^2 + y + 1 = 0\) also has no real solutions (discriminant \(< 0\)).

### Step 7: Conclusion about the curves
Since the curves do not intersect (as they do not have real solutions), we conclude that the two families of curves are orthogonal to each other.

### Final Answer
The two families of curves represented by the given equations are orthogonal.

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