If the slope of one of the lines represented by `ax^2 + 2hxy+by^2=0` be the nth power of the , prove that , `(ab^n)^(1/(n+1)) +(a^nb)^(1/(n+1))+2h=0`.

Answer

To solve the problem, we will follow these steps:

### Step 1: Start with the given equation
We are given the equation of the pair of straight lines:
\[ ax^2 + 2hxy + by^2 = 0 \]

### Step 2: Identify the slope
The slope \( m \) of the lines can be expressed as:
\[ m = \frac{y}{x} \]
We can rewrite the equation by dividing through by \( x^2 \):
\[ a + 2h \frac{y}{x} + b \left(\frac{y}{x}\right)^2 = 0 \]
Let \( m = \frac{y}{x} \), then:
\[ a + 2hm + bm^2 = 0 \]

### Step 3: Formulate the quadratic equation
This can be rearranged into a standard quadratic form:
\[ bm^2 + 2hm + a = 0 \]

### Step 4: Use the roots of the quadratic equation
Let the roots of the quadratic equation be \( \alpha \) and \( \alpha^n \) (where \( \alpha \) is the slope of one line and \( \alpha^n \) is the slope of the other line). By Vieta's formulas, we have:
1. Sum of roots:
\[ \alpha + \alpha^n = -\frac{2h}{b} \]
2. Product of roots:
\[ \alpha \cdot \alpha^n = \frac{a}{b} \]

### Step 5: Express the sum and product in terms of \( \alpha \)
From the sum of roots:
\[ \alpha + \alpha^n = -\frac{2h}{b} \]
From the product of roots:
\[ \alpha^{n+1} = \frac{a}{b} \]

### Step 6: Solve for \( \alpha \)
From the product equation, we can express \( \alpha \):
\[ \alpha = \left(\frac{a}{b}\right)^{\frac{1}{n+1}} \]

### Step 7: Substitute \( \alpha \) back into the sum equation
Substituting \( \alpha \) into the sum equation:
\[ \left(\frac{a}{b}\right)^{\frac{1}{n+1}} + \left(\frac{a}{b}\right)^{\frac{n}{n+1}} = -\frac{2h}{b} \]

### Step 8: Multiply through by \( b^{\frac{n+1}{n+1}} \)
Multiply through by \( b^{\frac{n+1}{n+1}} \):
\[ b^{\frac{n+1}{n+1}} \cdot \left(\frac{a}{b}\right)^{\frac{1}{n+1}} + b^{\frac{n+1}{n+1}} \cdot \left(\frac{a}{b}\right)^{\frac{n}{n+1}} = -2h \]

### Step 9: Simplify the equation
This gives us:
\[ a^{\frac{1}{n+1}} b^{-\frac{1}{n+1}} + a^{\frac{n}{n+1}} b^{-\frac{n}{n+1}} = -2h \]

### Step 10: Rearranging the equation
Rearranging gives us:
\[ (ab^n)^{\frac{1}{n+1}} + (a^n b)^{\frac{1}{n+1}} + 2h = 0 \]

Thus, we have proved that:
\[ (ab^n)^{\frac{1}{n+1}} + (a^n b)^{\frac{1}{n+1}} + 2h = 0 \]